CBSE Sample Paper 2023-24 · Section B · 2 marks

Q25: Does $f(x) = x^3 + x$ have any critical points?

Sample 2023-24Ch 6 · Application of Derivatives

The question

Check whether the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3 + x$ ⁠, has any critical point/s or not ? If yes, then find the point/s.

Exam answer

f(x) = x^3 + x, \text{ for all } x \in \mathbb{R}

f'(x) = 3x^2 + 1

Why?

A critical point is a point $c$ in the domain where

\begin{aligned} &f'(c) = 0, \text{ or} \\ &f'(c) \text{ does not exist} \end{aligned}

So we need $f'(x)$ first. By the power rule, $\frac{d}{dx}x^n = nx^{n - 1}$ ⁠, so $x^3$ gives $3x^2$ and $x$ gives 1.

For all $x \in \mathbb{R}$ ⁠, $x^2 \ge 0 \Rightarrow f'(x) \ge 1 > 0$ ⁠.

Why?

A square is never negative, so $3x^2 \ge 0$ ⁠. Adding 1 gives $3x^2 + 1 \ge 1$ ⁠.

So $f'(x) \ne 0$ for any $x$ ⁠, and $f'(x)$ exists for every $x$ ⁠.

Why?

Both ways of being critical fail. $f'(x)$ is at least 1, so it is never 0. And $f'(x)$ is a polynomial, so it exists at every real $x$ ⁠.

Watch out: $3x^2 + 1 = 0$ gives $x^2 = -\frac13$ ⁠, which has no real solution. There is no real critical point to report.

\begin{aligned} 3x^2 + 1 &= 0 \\ 3x^2 &= -1 \quad (\text{subtract } 1) \\ x^2 &= -\frac13 \quad (\text{divide by } 3) \end{aligned}

No real number squared is negative.

Hence, no critical point exists.

Where this comes from

Chapter 6, Application of Derivatives: Critical points of a function.

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Q25: Does f(x)=x3+xf(x) = x^3 + xf(x)=x3+x have any critical points?

Q25: Does $f(x) = x^3 + x$ have any critical points?