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

The least value of the function f(x) = ax + (b)/(x – 2nd PUC Mathematics

NCERT Exemplar · Q26

f(x) = ax + \dfrac{b}{x}

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Concept understanding — Rational Function Optimization

Rational Function Optimization

A rational function is a ratio of two polynomials, $f(x)=\dfrac{p(x)}{q(x)}$ — for example $f(x)=x+\dfrac1x$ or $f(x)=\dfrac{x^2+1}{x}$ . Finding its maximum or minimum is a common maxima–minima task, and the only new skill is differentiating a quotient cleanly.

The Method

To optimise $f(x)=\dfrac{p(x)}{q(x)}$ :

Differentiate with the quotient rule,

f'(x)=\frac{p'(x)\,q(x)-p(x)\,q'(x)}{\big(q(x)\big)^2}.

Set $f'(x)=0$ . A fraction is zero only when its numerator is zero, so you only need $p'q-pq'=0$ — the denominator never has to vanish.
Respect the domain. Values where $q(x)=0$ are excluded, and many problems restrict to $x>0$ . Keep these in mind when choosing which critical point is valid.
Classify each critical point with the second-derivative test or a sign check of $f'$ .

Tip

Before differentiating, simplify. Splitting $\dfrac{x^2+1}{x}=x+\dfrac1x$ turns an awkward quotient into an easy sum whose derivative is $1-\dfrac{1}{x^2}$ .

A Worked Example

Minimise $f(x)=x+\dfrac1x$ for $x>0$ .

Differentiating, $f'(x)=1-\dfrac{1}{x^2}$ . Setting this to zero gives $x^2=1$ , so $x=1$ (taking the positive root, since $x>0$ ). Then $f''(x)=\dfrac{2}{x^3}$ , and $f''(1)=2>0$ , confirming a minimum. The minimum value is $f(1)=1+1=2$ .

This matches the AM–GM bound $x+\dfrac1x\ge 2$ , with equality at $x=1$ — a useful sanity check.

Common Mistakes

Watch out

Setting the whole quotient's denominator to zero — you solve numerator $=0$ , not denominator $=0$ . …

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Setting $f'(x)=a-\dfrac{b}{x^2}=0$ gives $x=\sqrt{b/a}$ ; since $f''>0$ this is a minimum, and the least value is $2\sqrt{ab}$ .

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

f'(x)=a-\frac{b}{x^2}.

Step 2 — critical point

Set $f'(x)=0$ :

a-\frac{b}{x^2}=0\ \Rightarrow\ x^2=\frac{b}{a}\ \Rightarrow\ x=\sqrt{\frac{b}{a}}\quad(\text{positive root, since }x>0).

Step 3 — confirm it is a minimum

f''(x)=\frac{2b}{x^3}>0\quad\text{for }x>0,

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 $x=\sqrt{b/a}$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2024Set 2024-E1 markMCQ
$y=\frac{a x-b}{(x-1)(x-4)}$
›Reveal solutionSolution
The turning point at $P(2,-1)$ is a maximum because the second derivative is negative there, after using the given point to determine the constants $a$ and $b$ .

We are given the function
$y = \frac{ax - b}{(x-1)(x-4)}$
and told that $P(2,-1)$ is a turning point (i.e., a stationary point). That means two conditions hold at $x=2$ :

The curve passes through $(2,-1)$ , so $y(2) = -1$ .

The derivative $y'(2) = 0$ .

Our job is to classify this stationary point as a maximum, minimum, or neither.

1. Use the point to find a relation between $a$ and $b$

At $x=2$ :
$y(2) = \frac{2a - b}{(2-1)(2-4)} = \frac{2a - b}{(1)(-2)} = -\frac{2a - b}{2}.$
We know $y(2) = -1$ , so:
$-\frac{2a - b}{2} = -1 \quad\Rightarrow\quad \frac{2a - b}{2} = 1 \quad\Rightarrow\quad 2a - b = 2. \qquad(1)$

2. Differentiate and use the stationary condition

First, rewrite the denominator:
$y = \frac{ax - b}{(x-1)(x-4)} = \frac{ax - b}{x^2 - 5x + 4}.$
Differentiate using the quotient rule:
$y' = \frac{a(x^2 - 5x + 4) - (ax - b)(2x - 5)}{(x^2 - 5x + 4)^2}.$
At a stationary point, $y'(2) = 0$ , so the numerator must be zero at $x=2$ :
$a(4 - 10 + 4) - (2a - b)(4 - 5) = a(-2) - (2a - b)(-1) = -2a + (2a - b) = -b.$
Set equal to zero:
$-b = 0 \quad\Rightarrow\quad b = 0.$
Then from (1): $2a - 0 = 2 \Rightarrow a = 1$ .

So the function is:
$y = \frac{x}{(x-1)(x-4)}.$

3. Classify the stationary point using the second derivative

We could compute $y''$ directly, but it’s easier to use the sign of the second derivative at $x=2$ .

First, simplify $y'$ with $a=1, b=0$ :
$y' = \frac{(x^2 - 5x + 4) - x(2x - 5)}{(x^2 - 5x + 4)^2} = \frac{x^2 - 5x + 4 - 2x^2 + 5x}{(x^2 - 5x + 4)^2} = \frac{-x^2 + 4}{(x^2 - 5x + 4)^2}.$
So:
$y' = \frac{4 - x^2}{(x^2 - 5x + 4)^2}.$
Now differentiate again. Use the quotient rule on $y'$ :
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COMEDK 2021Set 20211 markMCQ
$f(x) = {1 \over {4{x^2} + 2x + 1}}$
›Reveal solutionSolution

Therefore the maximum value of f is 1/(3/4) = 4/3.

Concept: for f(x) = 1/g(x) with g(x) > 0 everywhere, f is maximum where g is minimum.

g(x) = 4x^2 + 2x + 1. Its discriminant is 4 - 16 = -12 < 0 and the leading coefficient is positive, so g(x) > 0 for all real x.

Minimum of g occurs at x = -b/(2a) = -2/(2*4) = -1/4. …
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Q.The least value of the function $f(x) = ax + \dfrac{b}{x}$ ( $a > 0$ , $b > 0$ , $x > 0$ ) is ______.

Concept understanding — Rational Function Optimization

Rational Function Optimization

The Method

A Worked Example

Common Mistakes

The idea

Step 1 — differentiate

Step 2 — critical point

Step 3 — confirm it is a minimum

Step 4 — the least value

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

Steps

Common Mistakes

1. Use the point to find a relation between $a$ and $b$

2. Differentiate and use the stationary condition

3. Classify the stationary point using the second derivative

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

The idea

Step 1 — differentiate

Step 2 — critical point

Step 3 — confirm it is a minimum

Step 4 — the least value

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

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