Q.Show that the function $f$ given by $f(x) = x^3 - 3x^2 + 4x,\ x \in \mathbb{R}$ is increasing on $\mathbb{R}$.

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If the function $y=g(x)$ representing the slopes of the tangents drawn to the curve $y=3x^4-5x^3-12x^2+18x+3$ is strictly increasing then the domain of $g(x)$ is (A) $\left[-\dfrac{1}{2},\dfrac{4}{3}\right]$ (B) $\left(\dfrac{-1}{2},\dfrac{4}{3}\right)$ (C) $\mathbb{R}-\left(\dfrac{-1}{2},\dfrac{3}{4}\right)$ (D) $\mathbb{R}-\left[\dfrac{-1}{2},\dfrac{4}{3}\right]$

   ›Reveal solutionSolution

   The "slope function" $g(x)$ is the derivative of the given quartic; its own strict increase is governed by $g'(x)>0$, i.e. a second derivative test producing a quadratic inequality. Answer: $\mathbb R-\left[-\frac12,\frac43\right]$.

   Concept and Intuition

   $g(x)$, "the slope of the tangent" to $y=3x^4-5x^3-12x^2+18x+3$, is exactly $y'(x)$ — a new function in its own right. Asking where this function is strictly increasing is asking where $g'(x)=y''(x)>0$, i.e. a standard increasing/decreasing analysis one derivative order up.

   Step-by-Step Solution

   1. $y=3x^4-5x^3-12x^2+18x+3$, so $g(x)=y'=12x^3-15x^2-24x+18$.
   2. $g$ is strictly increasing where $g'(x)>0$: $g'(x)=36x^2-30x-24$.
   3. Factor out 6: $g'(x)=6(6x^2-5x-4)$.
   4. Solve $6x^2-5x-4=0$: discriminant $=25+96=121=11^2$, so $x=\dfrac{5\pm11}{12}$, giving $x=\dfrac{16}{12}=\dfrac43$ and $x=\dfrac{-6}{12}=-\dfrac12$.
   5. Since the coefficient of $x^2$ is positive, $6x^2-5x-4>0$ outside the roots and $<0$ between them: positive for $x<-\dfrac12$ or $x>\dfrac43$.
   6. So $g'(x)>0 \iff x\in(-\infty,-\tfrac12)\cup(\tfrac43,\infty)$, i.e. the domain of strict increase of $g$ is $\mathbb R-\left[-\dfrac12,\dfrac43\right]$ (the closed interval between the roots is excluded, matching where $g'\le 0$).

   Common Mistakes

   • Confusing $g(x)$ (the slope function $y'$) with $y$ itself, and analyzing $y'>0$ instead of $g'(x)=y''>0$.
   • Sign/interval error — forgetting an upward parabola is positive outside, not between, its roots.

   ✓Final answer

   The correct option is (D) — $\mathbb{R}-\left[-\dfrac12,\dfrac43\right]$.

   ANSWER: D

2. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.For which value(s) of '$a$', $f(x) = -x^3 + 4ax^2 + 2x - 5$ is decreasing for every '$x$'? (A) $(1, 2)$ (B) $(3, 4)$ (C) $R$ (D) No value of '$a$'

   ›Reveal solutionSolution

   $f'(x)=-3x^2+8ax+2$ always attains a positive maximum for every real $a$, so $f$ can never be decreasing for all $x$ — the answer is "no value of $a$."

   Concept and Intuition

   For $f$ to be decreasing on all of $R$, its derivative must be $\le 0$ everywhere. $f'(x)$ here is a downward-opening parabola in $x$, so as $x\to\pm\infty$ it's automatically negative — the only risk is its peak (vertex) value going positive. If that peak is always positive regardless of $a$, no $a$ can work.

   Step-by-Step Solution

   1. Differentiate: $f'(x)=-3x^2+8ax+2$.
   2. This is a downward parabola in $x$ (leading coefficient $-3<0$), with vertex at $x_0=\dfrac{-8a}{2(-3)}=\dfrac{4a}{3}$.
   3. Maximum value of $f'$ at the vertex:

   $f'(x_0) = -3\left(\frac{4a}{3}\right)^2 + 8a\left(\frac{4a}{3}\right) + 2 = -\frac{16a^2}{3} + \frac{32a^2}{3} + 2 = \frac{16a^2}{3} + 2.$

   4. Since $\dfrac{16a^2}{3}\ge 0$ for every real $a$, the maximum of $f'$ is always $\ge 2 > 0$.
   5. Therefore $f'(x) > 0$ at least at the vertex for every value of $a$ — $f$ is never decreasing throughout $R$ for any $a$.

   Common Mistakes

   • Only checking $f'(x)\le 0$ at a couple of sample points instead of analyzing the vertex/discriminant of the quadratic.
   • Forgetting that a downward parabola can still have a positive region between its roots even though it goes to $-\infty$ at the ends.

   ✓Final answer

   The correct option is (D) — No value of '$a$'.

   ANSWER: D

3. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.In the interval $(-\infty, 0)$ the function $f(x) = x^2 + \dfrac{128}{x}$ (A) has only one local minimum value at $x = 4$ (B) has one maximum and one minimum at $x = 4$ and $x = -4$ respectively (C) is increasing (D) is decreasing

   ›Reveal solutionSolution

   Checking the sign of $f'(x)=2x-128/x^2$ on $(-\infty,0)$ shows it is always negative there, so the function is monotonically decreasing throughout the interval — no local max/min occurs in this domain.

   Concept and Intuition

   A function's monotonic behaviour on an interval is read off the sign of its derivative there. Critical points (where $f'=0$) only matter if they actually lie inside the interval in question; a critical point outside the interval is irrelevant to that interval's monotonicity.

   Step-by-Step Solution

   1. $f(x) = x^2 + \dfrac{128}{x}$. Differentiate: $f'(x) = 2x - \dfrac{128}{x^2}$.
   2. Find critical points: $f'(x)=0 \Rightarrow 2x = \dfrac{128}{x^2} \Rightarrow 2x^3 = 128 \Rightarrow x^3=64 \Rightarrow x=4$.
   3. The only critical point is $x=4$, which is not in $(-\infty,0)$.
   4. Check the sign of $f'(x)$ for any $x<0$: $2x$ is negative (since $x<0$); $x^2>0$ always, so $\dfrac{128}{x^2}>0$, making $-\dfrac{128}{x^2}$ negative.
   5. So $f'(x)$ is the sum of two negative quantities for every $x<0$: $f'(x)<0$ throughout $(-\infty,0)$.
   6. Since $f'(x)<0$ everywhere on this interval (no sign change, no zero), $f$ is strictly decreasing on all of $(-\infty,0)$ — there is no local max or min inside it.

   Common Mistakes

   • Solving $f'(x)=0$ and finding $x=4$, then wrongly assuming a mirror critical point at $x=-4$ exists too (it doesn't — check $f'(-4) = -8-8=-16\neq0$).
   • Assuming the interval must contain a turning point just because the full domain $\mathbb R\setminus\{0\}$ does.

   ✓Final answer

   The correct option is (D) — is decreasing.

   ANSWER: D

4. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.The maximum value of 'a' such that the second derivative of $x^4 + ax^3 + \dfrac{3x^2}{2} + 1$ is positive for all real $x$ is (A) $3$ (B) $-3$ (C) $2$ (D) $-2$

   ›Reveal solutionSolution

   The second derivative is a quadratic in $x$; positivity for all $x$ requires a non-positive discriminant, which bounds $a$.

   Concept and Intuition

   An upward-opening quadratic $Ax^2+Bx+C$ (here in $x$, with $A=12>0$) is $\ge0$ for all real $x$ exactly when its discriminant $B^2-4AC\le0$.

   Step-by-Step Solution

   1. $f(x)=x^4+ax^3+\frac{3x^2}{2}+1$.
   2. $f'(x)=4x^3+3ax^2+3x$.
   3. $f''(x)=12x^2+6ax+3$.
   4. Require $f''(x)\ge0$ for all real $x$ (boundary case of the required positivity): discriminant of $12x^2+6ax+3$ is $(6a)^2-4(12)(3)=36a^2-144$.
   5. Need $36a^2-144\le0\Rightarrow a^2\le4\Rightarrow -2\le a\le2$.
   6. The maximum value of $a$ satisfying this is $a=2$.

   Common Mistakes

   • Differentiating only once instead of twice.
   • Sign error in the discriminant formula, or forgetting the leading coefficient must be positive to conclude "always positive" from a non-positive discriminant.

   ✓Final answer

   The correct option is (C) — $2$.

   ANSWER: C

5. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If the tangent drawn to the curve $y=x^3-ax^2+x+1$ at each point $x\in\mathbb{R}$, is inclined at an acute angle with the positive direction of X-axis, then the set of all possible values of 'a' is (A) $\mathbb{R}-(-\sqrt3,\sqrt3)$ (B) $[-3,3]$ (C) $\mathbb{R}$ (D) $(-\sqrt3,\sqrt3)$

   ›Reveal solutionSolution

   The tangent slope $3x^2-2ax+1$ must stay strictly positive for all real $x$; requiring a negative discriminant gives $a\in(-\sqrt3,\sqrt3)$.

   Concept and Intuition

   "Tangent inclined at an acute angle with the positive x-axis" means the tangent's slope is strictly positive (an acute angle has $\tan\theta>0$). Since this must hold for every real $x$ (the curve's domain), the derivative — a quadratic in $x$ — must never touch or cross zero; it must be strictly positive throughout.

   Step-by-Step Solution

   1. $y = x^3-ax^2+x+1 \Rightarrow y' = 3x^2-2ax+1$.
   2. Require $y'>0$ for all $x\in\mathbb R$.
   3. A quadratic $Ax^2+Bx+C$ with $A>0$ is positive for all $x$ iff its discriminant $B^2-4AC<0$.
   4. Here $A=3,\ B=-2a,\ C=1$: discriminant $=4a^2-12$.
   5. Require $4a^2-12<0 \Rightarrow a^2<3 \Rightarrow -\sqrt3<a<\sqrt3$.

   Common Mistakes

   • Using $\le 0$ instead of $<0$ for the discriminant — equality would allow the slope to touch zero at one point (a horizontal tangent), which is not a strictly acute angle.
   • Forgetting to check the sign of the leading coefficient before applying the discriminant test.

   ✓Final answer

   The correct option is (D) — $(-\sqrt3,\sqrt3)$.

   ANSWER: D

6. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.Let $P(x)=x^4+ax^3+bx^2+cx+d$ be such that $x=0$ is the only real root of $P^1(x)=0$. If $P(-1)<P(1)$, then in the interval $[-1,1]$ (A) P(-1) is not minimum of P(x), but P(1) is the maximum of P(x) (B) P(-1) is minimum of P(x), but P(1) is not the maximum of P(x) (C) Neither P(-1) is the minimum nor P(1) is the maximum of P(x) (D) P(-1) is the minimum and P(1) is the maximum of P(x)

   ›Reveal solutionSolution

   This tests reading the sign of $P'$ from the structure of a cubic with a single real root; $P$ turns out to be strictly decreasing then increasing with its minimum at the interior point $x=0$, so $P(-1)$ is never the minimum, while the given inequality forces $P(1)$ to be the maximum.

   Concept and Intuition

   A quartic's monotonicity on an interval is governed by the sign of its derivative, a cubic here. If that cubic has only one real root, the other two roots are a complex-conjugate pair, so the cubic (as a real function) doesn't change sign there — it only changes sign at the single real root. That tells us $P$ has exactly one turning point on all of $\mathbb R$, at $x=0$, and it must be a minimum (since $P\to+\infty$ both ways, being a quartic with positive leading coefficient).

   Step-by-Step Solution

   1. $P'(x)=4x^3+3ax^2+2bx+c$. Since $x=0$ is a root, $P'(0)=c=0$.
   2. So $P'(x)=4x^3+3ax^2+2bx=x(4x^2+3ax+2b)$.
   3. For $x=0$ to be the only real root of $P'$, the quadratic factor $4x^2+3ax+2b$ must have no real zero, i.e. discriminant $9a^2-32b<0$. Since its leading coefficient $4>0$ and it has no real root, $4x^2+3ax+2b>0$ for all real $x$.
   4. Therefore $P'(x)=x\cdot(\text{always positive})$, so $P'(x)<0$ for $x<0$ and $P'(x)>0$ for $x>0$: $P$ is strictly decreasing on $[-1,0]$ and strictly increasing on $[0,1]$.
   5. On $[-1,1]$, the minimum of $P$ is therefore at the interior point $x=0$ (i.e. $P(0)<P(-1)$ and $P(0)<P(1)$), so $P(-1)$ is not the minimum.
   6. The maximum on $[-1,1]$ must be at one of the two endpoints. Since we're given $P(-1)<P(1)$, the maximum is $P(1)$.

   Common Mistakes

   • Assuming a single real root of $P'$ automatically makes $x=-1$ or $x=1$ extremal — the extremum is always at the root itself ($x=0$), not at the endpoints.
   • Concluding both $P(-1)$ is minimum and $P(1)$ is maximum just because they are endpoints, without checking $P(0)$ is actually lower than both.

   ✓Final answer

   The correct option is (A) — P(-1) is not minimum of P(x), but P(1) is the maximum of P(x).

   ANSWER: A

7. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If a number is drawn at random from the set $\{1, 3, 5, 7, \ldots, 59\}$, then the probability that it lies in the interval in which the function $f(x) = x^3 - 16x^2 + 20x - 5$ is strictly decreasing, is (A) $\dfrac{1}{5}$ (B) $\dfrac{1}{3}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{6}$

   ›Reveal solutionSolution

   Find where the cubic is decreasing (between the roots of its derivative), count the odd numbers from the set lying there, and divide by the set size; the answer is $\frac16$.

   Concept and Intuition

   A differentiable function is strictly decreasing exactly where its derivative is negative. For a cubic $f(x)=x^3-16x^2+20x-5$, $f'(x)$ is an upward-opening quadratic, so $f'(x)<0$ precisely between its two real roots. Once that interval is known, this becomes a plain classical-probability counting problem on a finite set.

   Step-by-Step Solution

   1. $f'(x) = 3x^2 - 32x + 20$.
   2. Solve $3x^2-32x+20=0$: discriminant $= 32^2-4\cdot3\cdot20 = 1024-240=784=28^2$.

   $x = \frac{32 \pm 28}{6} \implies x = 10 \text{ or } x = \frac{2}{3}.$

   3. Since the leading coefficient $3>0$, $f'(x)<0$ for $x\in\left(\frac23,10\right)$ — this is the decreasing interval.
   4. The set is $\{1,3,5,\dots,59\}$, the odd numbers from 1 to 59: total count $=\dfrac{59-1}{2}+1=30$.
   5. Members of the set lying strictly inside $\left(\frac23,10\right)$: $1,3,5,7,9$ — that's 5 numbers ($9<10$, and the next odd number $11>10$ is excluded).
   6. Required probability $=\dfrac{5}{30}=\dfrac{1}{6}$.

   Common Mistakes

   • Using the roots of $f(x)=0$ instead of $f'(x)=0$.
   • Including $x=10$ itself as "decreasing" (the boundary point is not strictly interior) — doesn't matter numerically here since 10 isn't odd, but conceptually the interval is open.
   • Miscounting the total elements of the arithmetic sequence.

   ✓Final answer

   The correct option is (D) — $\dfrac{1}{6}$.

   ANSWER: D

8. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If $f(x) = x^x$, then the interval in which $f(x)$ decreases is (A) $\left[0, \dfrac{1}{e}\right]$ (B) $[0, e]$ (C) $\left[\dfrac{1}{e}, \infty\right]$ (D) $[0, e^e]$

   ›Reveal solutionSolution

   Differentiate $x^x$ using logarithmic differentiation and find where the derivative is negative.

   Concept and Intuition

   A function decreases where its derivative is negative. Since $x^x$ itself is always positive on its domain $x>0$, the sign of $f'(x)$ is controlled entirely by the factor $(\log x+1)$.

   Step-by-Step Solution

   1. $f(x)=x^x$. Take logs: $\log f=x\log x$.
   2. Differentiate: $\dfrac{f'}{f}=\log x+1$, so $f'(x)=x^x(\log x+1)$.
   3. Since $x^x>0$ for all $x>0$, the sign of $f'(x)$ matches the sign of $(\log x+1)$.
   4. $f'(x)<0 \iff \log x+1<0 \iff \log x<-1 \iff x<e^{-1}=\dfrac1e$.
   5. So $f$ is decreasing on $\left(0,\dfrac1e\right)$, conventionally written as the closed interval $\left[0,\dfrac1e\right]$.

   Common Mistakes

   • Forgetting that $x^x$ requires $x>0$, and mistakenly extending the interval to negative $x$.
   • Sign error when solving $\log x+1<0$.

   ✓Final answer

   The correct option is (A) — $\left[0, \dfrac{1}{e}\right]$.

   ANSWER: A

9. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.At $x = 0$, $f(x) = \cos x - 1 + \dfrac{x^2}{2} - \dfrac{x^3}{3}$ (A) has a minimum value (B) has a maximum value (C) has no extremum value (D) is not defined

   ›Reveal solutionSolution

   The Taylor expansion of $f$ about $x=0$ has leading term $-x^3/3$ (odd power), so $f$ passes through $0$ changing sign — an inflection-type behavior, not an extremum.

   Concept and Intuition

   At a candidate critical point, if the first nonzero derivative is of odd order, the function does not have a local extremum there (it's increasing or decreasing straight through); only an even-order first-nonzero derivative gives a genuine min/max.

   Step-by-Step Solution

   1. $\cos x=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\dots$
   2. $f(x)=\cos x-1+\dfrac{x^2}{2}-\dfrac{x^3}{3}=\left(1-\dfrac{x^2}{2}+\dfrac{x^4}{24}\right)-1+\dfrac{x^2}{2}-\dfrac{x^3}{3}+\dots=-\dfrac{x^3}{3}+\dfrac{x^4}{24}+\dots$
   3. Near $x=0$, $f(x)\approx-\dfrac{x^3}{3}$: for small $x>0$, $f<0$; for small $x<0$, $f>0$ (since $-(-|x|)^3/3=|x|^3/3>0$).
   4. So $f$ changes sign through $x=0$ rather than staying one-signed on both sides — confirming this is not a local extremum.
   5. Cross-check via derivatives: $f'(0)=0$, $f''(0)=0$, but $f'''(0)=-2\ne0$ (odd order), consistent with "no extremum."

   Common Mistakes

   • Stopping at $f'(0)=f''(0)=0$ and wrongly concluding it's a minimum without checking the sign behavior or higher derivatives.
   • Sign errors in the Taylor expansion cancelling the $x^2$ terms.

   ✓Final answer

   The correct option is (C) — has no extremum value.

   ANSWER: C

10. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.The cubic equation $2x^3 - 3x^2 + 6x + 2 = 0$ (A) has 3 distinct real roots (B) has only one real root in the interval $(-1, 0)$ (C) has two distinct real roots (D) has only one real root in the interval $(0, 1)$

    ›Reveal solutionSolution

    Showing $f'(x)>0$ always (via a negative-discriminant quadratic) proves the cubic is strictly increasing and has exactly one real root; sign checking at $-1$ and $0$ locates it in $(-1,0)$.

    Concept and Intuition

    A cubic with a strictly positive (or always negative) derivative is monotonic, and a monotonic continuous function can cross zero at most once — so it has exactly one real root. To locate that root, use the Intermediate Value Theorem: find two points where the function has opposite signs, and the root must lie between them.

    Step-by-Step Solution

    1. Let $f(x)=2x^3-3x^2+6x+2$. Compute $f'(x) = 6x^2-6x+6 = 6(x^2-x+1)$.
    2. Check the discriminant of $x^2-x+1$: $D = (-1)^2 - 4(1)(1) = 1-4=-3 <0$. Since the discriminant is negative and the leading coefficient is positive, $x^2-x+1>0$ for all real $x$.
    3. So $f'(x) = 6(x^2-x+1) > 0$ for all $x$ — $f$ is strictly increasing on $\mathbb{R}$, hence it can have at most (and, being an odd-degree polynomial, exactly) one real root.
    4. This immediately rules out "3 distinct real roots" and "two distinct real roots".
    5. To locate the root, evaluate $f$ at convenient points: $f(-1) = 2(-1)-3(1)+6(-1)+2 = -2-3-6+2=-9$ (negative). $f(0) = 0-0+0+2=2$ (positive). $f(1)=2-3+6+2=7$ (positive).
    6. Since $f(-1)<0$ and $f(0)>0$, by the Intermediate Value Theorem the single real root lies in $(-1,0)$. Since $f(0)$ and $f(1)$ are both positive, there's no root in $(0,1)$.

    Common Mistakes

    • Assuming a cubic must have 3 real roots without checking the derivative's sign (a cubic can have just 1 real root and a complex conjugate pair).
    • Checking the wrong interval or making an arithmetic sign error while evaluating $f$ at test points.

    ✓Final answer

    The correct option is (B) — has only one real root in the interval $(-1, 0)$.

    ANSWER: B

11. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.$A(1,15)$, $B(3,-12)$, $C(6,12)$ are three consecutive turning points of a continuous curve $y = f(x)$. If $f(x) = 0$ only for $x = \alpha$ and $x = \beta$, then $|\beta - \alpha| <$ (A) $27$ (B) $2$ (C) $5$ (D) $24$

    ›Reveal solutionSolution

    The two zeros lie strictly between consecutive turning points; bounding $\alpha\in(1,3)$ and $\beta\in(3,6)$ gives $|\beta-\alpha|<5$.

    Concept and Intuition

    Between a positive local extreme value and a negative one, a continuous curve must cross zero at least once (Intermediate Value Theorem). The turning points bracket where each zero can occur.

    Step-by-Step Solution

    1. At $x=1$, $f=15>0$ (local extreme); at $x=3$, $f=-12<0$ (local extreme); so by IVT, $f$ has a zero $\alpha$ strictly between $1$ and $3$: $1<\alpha<3$.
    2. At $x=3$, $f=-12<0$; at $x=6$, $f=12>0$; so $f$ has a zero $\beta$ strictly between $3$ and $6$: $3<\beta<6$.
    3. Since these are the only two zeros given, $\beta-\alpha$ is bounded: the largest possible value approaches $6-1=5$ (but is never reached since both bounds are open), so $|\beta-\alpha|<5$.

    Common Mistakes

    • Using closed rather than open bounds, which could make it seem like equality (=5) is achievable.

    ✓Final answer

    The correct option is (C) — $5$.

    ANSWER: C

12. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.The curve represented by $x=t^5+5t^3+20t+7$ and $y=4t^3-3t^2-18t+3$ is decreasing in the interval (A) $(-2,-1)$ (B) $(3/2,2)$ (C) $(-1,3/2)$ (D) $(-2,2)$

    ›Reveal solutionSolution

    For a parametric curve with $x'(t)$ always positive, the curve decreases in $y$ exactly where $y'(t)<0$. Here that interval is $(-1,3/2)$.

    Concept and Intuition

    For a curve given parametrically, $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$. The curve is 'decreasing' (as a function $y$ of $x$) precisely where this ratio is negative. If $dx/dt$ never changes sign (stays positive throughout), then the sign of $dy/dx$ is simply the sign of $dy/dt$ — so we only need to analyze $dy/dt$.

    Step-by-Step Solution

    1. Differentiate $x=t^5+5t^3+20t+7$: $\dfrac{dx}{dt}=5t^4+15t^2+20=5(t^4+3t^2+4)$.
    2. Check the sign of $t^4+3t^2+4$: substituting $u=t^2\ge0$, this is $u^2+3u+4$, whose discriminant is $9-16=-7<0$, so it's always positive. Hence $\dfrac{dx}{dt}>0$ for every real $t$ — $x$ is strictly increasing in $t$.
    3. Differentiate $y=4t^3-3t^2-18t+3$: $\dfrac{dy}{dt}=12t^2-6t-18=6(2t^2-t-3)=6(2t-3)(t+1)$.
    4. Since $dx/dt>0$ always, the curve decreases (in $y$ vs $x$) exactly when $dy/dt<0$, i.e. when $(2t-3)(t+1)<0$.
    5. The roots are $t=-1$ and $t=3/2$; the product of two linear factors is negative strictly between the roots: $-1<t<3/2$.

    Common Mistakes

    • Forgetting to check that $dx/dt$ doesn't change sign — if it did, you'd need to track both signs together rather than just $dy/dt$.
    • Sign errors factoring the quadratic $2t^2-t-3=(2t-3)(t+1)$.

    ✓Final answer

    The correct option is (C) — $(-1,3/2)$.

    ANSWER: C