Q.Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

We use the first derivative test: find $f'(x)$, set it to zero for critical points, then check the sign change of $f'(x)$ around each point to classify maxima/minima.

(i) $f(x)=x^2$

$f'(x)=2x=0 \Rightarrow x=0$. For $x<0$, $f'(x)<0$; for $x>0$, $f'(x)>0$. So $x=0$ is a local minimum.

$f(0)=0$.

(ii) $g(x)=x^3-3x$

$g'(x)=3x^2-3=3(x-1)(x+1)=0 \Rightarrow x=-1,1$.

Sign of $g'(x)$: $+$ for $x<-1$, $-$ for $-1<x<1$, $+$ for $x>1$.

So $x=-1$ is a local max ($g(-1)=2$), $x=1$ is a local min ($g(1)=-2$).

(iii) $h(x)=\sin x+\cos x$, $0<x<\frac{\pi}{2}$

$h'(x)=\cos x-\sin x=0 \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}$.

For $x<\frac{\pi}{4}$, $h'(x)>0$; for $x>\frac{\pi}{4}$, $h'(x)<0$. So local max.

$h(\frac{\pi}{4})=\sqrt{2}$.

(iv) $f(x)=\sin x-\cos x$, $0<x<2\pi$

$f'(x)=\cos x+\sin x=0 \Rightarrow \tan x=-1 \Rightarrow x=\frac{3\pi}{4},\frac{7\pi}{4}$.

Sign of $f'(x)$: $+$ for $0<x<\frac{3\pi}{4}$, $-$ for $\frac{3\pi}{4}<x<\frac{7\pi}{4}$, $+$ for $\frac{7\pi}{4}<x<2\pi$.

So $x=\frac{3\pi}{4}$ is local max ($f=\sqrt{2}$), $x=\frac{7\pi}{4}$ is local min ($f=-\sqrt{2}$).

(v) $f(x)=x^3-6x^2+9x+15$

$f'(x)=3x^2-12x+9=3(x-1)(x-3)=0 \Rightarrow x=1,3$.

Sign: $+$ for $x<1$, $-$ for $1<x<3$, $+$ for $x>3$.

So $x=1$ local max ($f(1)=19$), $x=3$ local min ($f(3)=15$).

(vi) $g(x)=\frac{x}{2}+\frac{2}{x}$, $x>0$

$g'(x)=\frac12-\frac{2}{x^2}=0 \Rightarrow x^2=4 \Rightarrow x=2$ (positive root).

For $0<x<2$, $g'(x)<0$; for $x>2$, $g'(x)>0$. So local min.

$g(2)=2$.

(vii) $g(x)=\frac{1}{x^2+2}$

$g'(x)=-\frac{2x}{(x^2+2)^2}=0 \Rightarrow x=0$.

For $x<0$, $g'(x)>0$; for $x>0$, $g'(x)<0$. So local max.

$g(0)=\frac12$.

(viii) $f(x)=x\sqrt{1-x}$, $0<x<1$

$f'(x)=\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}=\frac{2(1-x)-x}{2\sqrt{1-x}}=\frac{2-3x}{2\sqrt{1-x}}=0 \Rightarrow x=\frac23$.

For $0<x<\frac23$, $f'(x)>0$; for $\frac23<x<1$, $f'(x)<0$. So local max.

$f(\frac23)=\frac23\sqrt{\frac13}=\frac{2}{3\sqrt3}$.

The core idea: the derivative $f'(x)$ tells us where a function is increasing ($f'>0$) or decreasing ($f'<0$). At a point where $f'$ changes sign — from positive to negative (max) or negative to positive (min) — we have a local extremum. If $f'$ does not change sign, the point is neither a max nor a min (e.g., an inflection point). This is the First Derivative Test.

Let’s go through each function step by step.

(i) $f(x) = x^2$

1. Derivative: $f'(x) = 2x$. Set $f'(x)=0 \Rightarrow 2x=0 \Rightarrow x=0$. This is the only critical point.
2. Sign analysis: For $x<0$, $f'(x)=2x<0$ (decreasing). For $x>0$, $f'(x)>0$ (increasing). So $f'$ changes from negative to positive at $x=0$.
3. Conclusion: Local minimum at $x=0$. Value: $f(0)=0$.

(ii) $g(x) = x^3 - 3x$

1. Derivative: $g'(x) = 3x^2 - 3 = 3(x^2-1) = 3(x-1)(x+1)$. Critical points: $x=1$ and $x=-1$.
2. Sign analysis: Test intervals around $-1$ and $1$.
   • For $x<-1$, say $x=-2$: $g'(-2)=3(4-1)=9>0$ (increasing).
   • Between $-1$ and $1$, say $x=0$: $g'(0)=-3<0$ (decreasing).
   • For $x>1$, say $x=2$: $g'(2)=3(4-1)=9>0$ (increasing). So at $x=-1$, $g'$ changes from positive to negative → local maximum. At $x=1$, $g'$ changes from negative to positive → local minimum.
3. Values: $g(-1)=(-1)^3 - 3(-1) = -1+3=2$. $g(1)=1-3=-2$.

(iii) $h(x) = \sin x + \cos x$, $0 < x < \frac{\pi}{2}$

1. Derivative: $h'(x) = \cos x - \sin x$. Set $h'(x)=0 \Rightarrow \cos x = \sin x \Rightarrow \tan x = 1$. In $(0,\pi/2)$, $x=\pi/4$.
2. Sign analysis: For $x<\pi/4$, say $x=0$: $h'(0)=1-0=1>0$ (increasing). For $x>\pi/4$, say $x=\pi/3$: $h'(\pi/3)=\cos(\pi/3)-\sin(\pi/3)=0.5 - 0.866 <0$ (decreasing). So $h'$ changes from positive to negative → local maximum.
3. Value: $h(\pi/4)=\sin(\pi/4)+\cos(\pi/4)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$.

(iv) $f(x) = \sin x - \cos x$, $0 < x < 2\pi$

1. Derivative: $f'(x) = \cos x + \sin x$. Set $f'(x)=0 \Rightarrow \cos x = -\sin x \Rightarrow \tan x = -1$. In $(0,2\pi)$, solutions: $x=3\pi/4$ and $x=7\pi/4$.
2. Sign analysis: Test intervals.
   • For $x$ just less than $3\pi/4$, say $x=\pi/2$: $f'(\pi/2)=0+1=1>0$ (increasing).
   • Between $3\pi/4$ and $7\pi/4$, say $x=\pi$: $f'(\pi)=-1+0=-1<0$ (decreasing).
   • For $x>7\pi/4$, say $x=2\pi$ (but note $2\pi$ is not included, so take $x=3\pi/2$? Actually $3\pi/2$ is less than $7\pi/4$? Wait: $3\pi/2 = 6\pi/4 = 1.5\pi$, and $7\pi/4=1.75\pi$, so $3\pi/2$ is between them. Let's pick $x=5\pi/3$ which is $>7\pi/4$? $5\pi/3 \approx 5.236$, $7\pi/4 \approx 5.498$, so $5\pi/3 < 7\pi/4$. Better: take $x=2\pi$ (not allowed) but we can take $x=11\pi/6 \approx 5.76$ which is $>7\pi/4$: $f'(11\pi/6)=\cos(11\pi/6)+\sin(11\pi/6)=\sqrt{3}/2 + (-1/2) \approx 0.866-0.5=0.366>0$ (increasing). So at $x=3\pi/4$, $f'$ changes from positive to negative → local maximum. At $x=7\pi/4$, $f'$ changes from negative to positive → local minimum.
3. Values: $f(3\pi/4)=\sin(3\pi/4)-\cos(3\pi/4)=\frac{\sqrt{2}}{2} - (-\frac{\sqrt{2}}{2}) = \sqrt{2}$. $f(7\pi/4)=\sin(7\pi/4)-\cos(7\pi/4)=-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}$.

(v) $f(x) = x^3 - 6x^2 + 9x + 15$

1. Derivative: $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$. Critical points: $x=1$, $x=3$.
2. Sign analysis:
   • For $x<1$, say $x=0$: $f'(0)=9>0$ (increasing).
   • Between $1$ and $3$, say $x=2$: $f'(2)=3(2-1)(2-3)=3(1)(-1)=-3<0$ (decreasing).
   • For $x>3$, say $x=4$: $f'(4)=3(3)(1)=9>0$ (increasing). So at $x=1$, $f'$ changes from positive to negative → local maximum. At $x=3$, $f'$ changes from negative to positive → local minimum.
3. Values: $f(1)=1-6+9+15=19$. $f(3)=27-54+27+15=15$.

(vi) $g(x) = \frac{x}{2} + \frac{2}{x}$, $x > 0$

1. Derivative: $g'(x) = \frac{1}{2} - \frac{2}{x^2}$. Set $g'(x)=0 \Rightarrow \frac{1}{2} = \frac{2}{x^2} \Rightarrow x^2 = 4 \Rightarrow x=2$ (since $x>0$).
2. Sign analysis: For $0<x<2$, say $x=1$: $g'(1)=0.5 - 2 = -1.5<0$ (decreasing). For $x>2$, say $x=4$: $g'(4)=0.5 - 2/16 = 0.5 - 0.125 = 0.375>0$ (increasing). So $g'$ changes from negative to positive → local minimum.
3. Value: $g(2)= \frac{2}{2} + \frac{2}{2} = 1+1=2$.

(vii) $g(x) = \frac{1}{x^2+2}$

1. Derivative: $g'(x) = -\frac{2x}{(x^2+2)^2}$. Set $g'(x)=0 \Rightarrow -2x=0 \Rightarrow x=0$. (Denominator never zero.)
2. Sign analysis: For $x<0$, say $x=-1$: $g'(-1)= -\frac{2(-1)}{(1+2)^2} = \frac{2}{9}>0$ (increasing). For $x>0$, say $x=1$: $g'(1)= -\frac{2}{(1+2)^2} = -\frac{2}{9}<0$ (decreasing). So $g'$ changes from positive to negative → local maximum.
3. Value: $g(0)=\frac{1}{0+2}=\frac{1}{2}$.

(viii) $f(x) = x\sqrt{1-x}$, $0 < x < 1$

1. Derivative: Write $f(x)=x(1-x)^{1/2}$. Use product rule: $f'(x) = (1)(1-x)^{1/2} + x \cdot \frac{1}{2}(1-x)^{-1/2}(-1) = \sqrt{1-x} - \frac{x}{2\sqrt{1-x}}$. Combine: $f'(x) = \frac{2(1-x) - x}{2\sqrt{1-x}} = \frac{2-2x - x}{2\sqrt{1-x}} = \frac{2-3x}{2\sqrt{1-x}}$. Set $f'(x)=0 \Rightarrow 2-3x=0 \Rightarrow x=\frac{2}{3}$. (Denominator is positive for $0<x<1$, so no other critical points.)
2. Sign analysis: For $x<2/3$, say $x=0.5$: $f'(0.5)=\frac{2-1.5}{2\sqrt{0.5}} = \frac{0.5}{2\cdot 0.707} >0$ (increasing). For $x>2/3$, say $x=0.8$: $f'(0.8)=\frac{2-2.4}{2\sqrt{0.2}} = \frac{-0.4}{2\cdot 0.447} <0$ (decreasing). So $f'$ changes from positive to negative → local maximum.
3. Value: $f(2/3) = \frac{2}{3}\sqrt{1-\frac{2}{3}} = \frac{2}{3}\sqrt{\frac{1}{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}$.

Method: The First Derivative Test for Local Maxima and Minima

This is the general procedure for locating and classifying every local turning point of a differentiable function, and it is what parts (i)-(viii) all use, whatever the function's shape.

Steps

Step 1: Differentiate the function.

Find $f'(x)$ using the standard differentiation rules (power, product, quotient, chain) as needed.

Step 2: Solve $f'(x) = 0$ to get the candidate critical points, restricted to the given domain.

Every candidate must lie inside the interval the question specifies (an open interval like $0<x<\frac{\pi}{2}$, or a restriction like $x>0$) — discard any solution that falls outside it.

Step 3: Test the sign of $f'(x)$ just before and just after each critical point.

Step 4: Substitute the classified critical point back into $f(x)$ to state the actual local maximum/minimum value.

The critical point itself is only the $x$-location; the question always also wants the function's value there.

Step 5 (Applying to this problem): Repeat Steps 1-4 independently for each of the eight sub-parts.

Each part has its own derivative and its own domain restriction — a critical point valid for one part (e.g. $x=-1$ for a cubic on all of $\mathbb{R}$) may be meaningless for another part restricted to $x>0$, so never carry a critical point across sub-parts.

Common Mistakes

Mistake 1: Treating every zero of $f'(x)$ as an automatic extremum.

Why it's wrong: a zero derivative only flags a candidate — if $f'$ keeps the same sign on both sides (as it can for functions related to $x^3$), the point is a saddle, not a maximum or minimum. Correct approach: always test the sign of $f'$ on both sides of the candidate before classifying it.

Mistake 2: Accepting a critical point that lies outside the question's stated domain.

Why it's wrong: in part (vi), $g'(x)=0$ mathematically also has the solution $x=-2$, but the question restricts $x>0$, so $x=-2$ is not a valid critical point here — including it would wrongly suggest a second turning point. Correct approach: solve $f'(x)=0$ freely, then immediately discard any root outside the given interval before doing the sign test.

Mistake 3: Mishandling the quotient/product-rule algebra in parts (vi)-(viii) and losing a sign.

Why it's wrong: expressions like $\frac{1}{2}-\frac{2}{x^2}$ or $\frac{2-3x}{2\sqrt{1-x}}$ are easy to differentiate incorrectly, which flips which side is increasing and which is decreasing — misclassifying a maximum as a minimum. Correct approach: re-derive the sign of the numerator and denominator separately before combining, and sanity-check with one concrete test value.