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

Key idea: Minimise $f(x)=ax+\dfrac{b}{x}$ ($x>0$) using calculus (or AM–GM).

Set $f'(x)=0$: $a=\dfrac{b}{x^2}\Rightarrow x^2=\dfrac{b}{a}\Rightarrow x=\sqrt{\tfrac{b}{a}}$ (positive root).

Since $f''(x)=\dfrac{2b}{x^3}>0$ for $x>0$, this is a minimum. Substituting: …

The idea

We want the smallest value of $f(x)=ax+\dfrac{b}{x}$ for $x>0$, with $a,b>0$. As $x\to0^+$ the term $\dfrac{b}{x}\to+\infty$, and as $x\to\infty$ the term $ax\to+\infty$, so somewhere in between the sum bottoms out. We find that turning point with the derivative.

Step 1 — differentiate

Step 2 — critical point

Set $f'(x)=0$:

Step 3 — confirm it is a minimum

so the function is concave up and the critical point is a minimum (and it is the global minimum, as $f\to+\infty$ at both ends).

Step 4 — the least value

Substitute $x=\sqrt{b/a}$: …

Method: Finding Extrema of Functions of the Form $ax + \dfrac{b}{x}$

This shape — a positive multiple of $x$ plus a positive multiple of $1/x$ — recurs constantly (cost functions, quantities that trade off against each other as $x$ grows), and always has exactly one extremum on $x>0$, found the same way every time.

Steps

Step 1: Differentiate.

Step 2: Set $f'(x)=0$ and solve for the critical point.

keeping only the positive root when the domain requires $x>0$.

Step 3: Classify the critical point with the second derivative.

For $x>0$ (and $b>0$), $f''(x)>0$ always — so the single critical point is always a minimum on this domain (never a maximum), and since $f(x)\to\infty$ at both ends of $(0,\infty)$, it is automatically the global minimum too.

Step 4 (Applying to this problem): substitute back to get the extreme value. …

Common Mistakes

Mistake 1: Keeping the negative root $x=-\sqrt{b/a}$

Why it's wrong: Solving $x^2=b/a$ algebraically gives $x=\pm\sqrt{b/a}$, but the problem restricts $x>0$ — carrying the negative root forward (or averaging both) gives a nonsensical or wrong critical point. Correct approach: discard the negative root immediately since it falls outside the given domain.

Mistake 2: Assuming the critical point is a minimum without checking …