Q.The least value of the function $f(x) = ax + \dfrac{b}{x}$ ($a > 0$, $b > 0$, $x > 0$) is ______.

Concept and Intuition

When numerator and denominator share the same quadratic core, subtract to expose the shared part: $\dfrac{u+10}{u}=1+\dfrac{10}{u}$. Since $10/u$ is a decreasing function of $u$ for $u>0$, maximizing the whole expression reduces to minimizing the quadratic denominator $u$.

Step-by-Step Solution

1. Let $u=3x^2+9x+7$. Then $3x^2+9x+17=u+10$, so the expression is $\dfrac{u+10}{u}=1+\dfrac{10}{u}$.
2. Find the range of $u$ for real $x$: $u$ is an upward-opening parabola (coefficient $3>0$), minimized at $x=-\dfrac{9}{2(3)}=-\dfrac32$.
3. $u_{\min}=3\left(\dfrac94\right)+9\left(-\dfrac32\right)+7=\dfrac{27}{4}-\dfrac{27}{2}+7=\dfrac{27-54+28}{4}=\dfrac14$.
4. So $u$ ranges over $\left[\dfrac14,\infty\right)$, always positive — the original fraction is defined for all real $x$. …

We want the maximum and minimum values of

for real $x$. The denominator is always positive (its discriminant $1-4 = -3 < 0$), so $y$ is defined for all real $x$.

Why this approach works:

Instead of calculus, we treat $y$ as a parameter and ask: for which $y$ does the equation have a real solution $x$? That turns the problem into a quadratic in $x$, and the condition $\text{discriminant} \ge 0$ gives the range of $y$. This is the classic method for rational functions with quadratic numerator and denominator.

1. Cross-multiply and rearrange

Bring all terms to one side:

Group powers of $x$:

2. Consider the case $y = 1$ separately

   If $y = 1$, the equation becomes $0 \cdot x^2 + 2x + 0 = 0$, so $x = 0$. Thus $y = 1$ is attainable. This will be an interior point of the range, not an extremum.

3. For $y \neq 1$, we have a quadratic in $x$

   Real $x$ exists iff the discriminant is non‑negative:

Simplify:

Expand:

Multiply by $-1$ (reversing inequality):

4. Solve the quadratic inequality Factor: $3y^2 - 10y + 3 = (3y - 1)(y - 3)$ …

Concept and Intuition

For $y=f(x)$ where $f$ is a ratio of two quadratics, cross-multiplying gives a quadratic in $x$ whose coefficients depend on $y$. Real $x$ exists only where its discriminant is $\ge 0$.

Step-by-Step Solution

1. Let $y = \dfrac{x^2-2x+3}{x^2-4x+7}$. Denominator $=(x-2)^2+3>0$ always.
2. Cross-multiply: $y(x^2-4x+7) = x^2-2x+3$.
3. Rearrange: $(y-1)x^2 + (-4y+2)x + (7y-3) = 0$.
4. Discriminant $\ge 0$: $(-4y+2)^2 - 4(y-1)(7y-3) \ge 0$.
5. Expand: $16y^2-16y+4-28y^2+40y-12\ge 0 \Rightarrow -12y^2+24y-8\ge 0$.
6. Multiply by $-1$: $12y^2-24y+8 \le 0$, divide by 4: $3y^2-6y+2\le 0$. …

Concept and Intuition

For a rational function $y=\dfrac{ax^2+bx+c}{px^2+qx+r}$ that must equal a value $y$ for SOME real $x$, cross-multiplying gives a quadratic in $x$ whose discriminant must be $\ge0$ for a real solution to exist. Solving this discriminant inequality for $y$ gives the full range of the function, and its lower bound is the minimum.

Step-by-Step Solution

1. Let $y=\dfrac{x^2+2x+5}{x^2+4x+10}$. Cross-multiply: $y(x^2+4x+10) = x^2+2x+5$.
2. Rearrange: $(y-1)x^2 + (4y-2)x + (10y-5) = 0$.
3. For real $x$ to exist, discriminant $\ge0$: $(4y-2)^2 - 4(y-1)(10y-5)\ge0$.
4. Expand: $(16y^2-16y+4) - (40y^2-60y+20) \ge 0 \Rightarrow -24y^2+44y-16\ge0$.
5. Divide by $-4$ (flip inequality): $6y^2-11y+4\le0$.
6. Solve $6y^2-11y+4=0$: $y=\dfrac{11\pm\sqrt{121-96}}{12}=\dfrac{11\pm5}{12}$, giving $y=\tfrac43$ or $y=\tfrac12$. …

Concept and Intuition

Whenever a rational function looks like $\dfrac{x}{\text{quadratic in }x}$, dividing numerator and denominator by $x$ (for $x\neq0$) often turns it into a reciprocal of a sum like $x+1/x$, which AM-GM handles beautifully: $x+\dfrac1x\ge2$ for $x>0$, with equality exactly at $x=1$.

Step-by-Step Solution

1. For $x>0$, divide numerator and denominator of $f(x)=\dfrac{x}{x^2+4x+1}$ by $x$: $f(x) = \dfrac{1}{x+4+\frac1x}$.
2. By AM-GM, for $x>0$: $x+\dfrac1x \ge 2\sqrt{x\cdot\frac1x} = 2$, with equality iff $x=1/x$, i.e. $x=1$.
3. So the denominator $x+4+\dfrac1x \ge 2+4=6$.
4. Therefore $f(x) = \dfrac{1}{x+4+1/x} \le \dfrac16$. …

Concept and Intuition

Expressions of the form $t+\dfrac kt$ (for $t>0$, $k>0$) have a well-known minimum value $2\sqrt k$ by the AM-GM inequality, attained when $t=\sqrt k$ — recognizing this shape after a simple substitution avoids full calculus.

Step-by-Step Solution

1. Let $t=x+2$ (considering the branch $x>-2$, i.e. $t>0$, where the function attains a genuine local/global minimum on that domain).
2. $f(x) = x+\dfrac{4}{x+2} = (t-2)+\dfrac4t = t+\dfrac4t-2$.
3. By AM-GM: for $t>0$, $t+\dfrac4t \ge 2\sqrt{t\cdot\frac4t}=2\sqrt4=4$, with equality when $t=\dfrac4t \Rightarrow t^2=4\Rightarrow t=2$ (taking the positive root). …

Concept and Intuition

The combination $x^2+1/x^2 = (x-1/x)^2+2$ is a standard substitution trick, turning a messy rational function of $x$ into a clean function of a single new variable $t=x-1/x$, which ranges over all nonzero reals as $x$ ranges over $\mathbb R-\{-1,0,1\}$.

Step-by-Step Solution

1. Let $t=g(x)=x-\dfrac1x$. Then $t^2=x^2-2+\dfrac{1}{x^2}$, so $f(x)=x^2+\dfrac1{x^2}=t^2+2$.
2. So $\dfrac{f(x)}{g(x)} = \dfrac{t^2+2}{t} = t+\dfrac2t =: h(t)$, where $t$ ranges over all real numbers except $0$.
3. Differentiate: $h'(t)=1-\dfrac{2}{t^2}$. Setting $h'(t)=0$ gives $t^2=2\Rightarrow t=\pm\sqrt2$.
4. $h''(t)=\dfrac{4}{t^3}$. At $t=\sqrt2$, $h''>0$, so this is a local minimum; at $t=-\sqrt2$, $h''<0$, a local maximum. …

Concept and Intuition

Whenever an expression has a term and its reciprocal (both positive), AM-GM instantly gives the minimum of their sum, achieved when the two terms are equal.

Step-by-Step Solution

1. Substitute $\sec^2\alpha = 1+\tan^2\alpha$: $5\tan^2\alpha+\dfrac{9}{\tan^2\alpha}+4(1+\tan^2\alpha) = 5\tan^2\alpha+4\tan^2\alpha+\dfrac{9}{\tan^2\alpha}+4 = 9\tan^2\alpha+\dfrac{9}{\tan^2\alpha}+4$.
2. Let $t=\tan^2\alpha>0$. Expression $=9t+\dfrac9t+4$.
3. By AM-GM: $9t+\dfrac9t \ge 2\sqrt{9t\cdot\frac9t}=2\sqrt{81}=18$, with equality when $9t=\dfrac9t\Rightarrow t=1$ (i.e. $\tan^2\alpha=1$, achievable). …