UPSC 2021 Maths Optional Paper 2 Q2a — Step-by-Step Solution

15 marks · Section A

Maxima and minima of single-variable functions · Real Analysis · asked 3× in 13 yrs · Read the full method →

Question

Find max and min of $f(x)=x^3-9x^2+26x-24$ for $0\le x\le 1$ .

Critical points via $f'=0$ ; check whether inside the interval; if not, monotone on the interval; max/min at endpoints.

$f'(x)=3x^2-18x+26$ .

$f'(x)=0\Rightarrow x=\dfrac{18\pm\sqrt{324-312}}{6}=\dfrac{18\pm\sqrt{12}}{6}=3\pm\dfrac{\sqrt 3}{3}=3\pm\dfrac{1}{\sqrt 3}$ .

Numerical: $1/\sqrt 3\approx 0.577$ . So critical points at $x\approx 2.423$ and $x\approx 3.577$ . Both outside $[0,1]$ .

$f'(0)=0-0+26=26>0$ . $f'(1)=3-18+26=11>0$ .

$f'$ positive on $[0,1]$ (no roots there) ⇒ $f$ strictly increasing on $[0,1]$ .

So:

\boxed{\;f_{\max}=-6\text{ at }x=1;\;\;f_{\min}=-24\text{ at }x=0.\;}