The value of the function f(x) = (x - 1)(x - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x) = (x - 1)(x - 2)^2$. Expanding the function: $f(x) = (x - 1)(x^2 - 4x + 4) = x^3 - 5x^2 + 8x - 4$. To find the critical points,we calc

Vedclass · Accepted Answer

$\frac{4}{27}$

Vedclass · Answer

(D) Given $f(x) = (x - 1)(x - 2)^2$. Expanding the function: $f(x) = (x - 1)(x^2 - 4x + 4) = x^3 - 5x^2 + 8x - 4$. To find the critical points,we calculate the first derivative: $f'(x) = 3x^2 - 10x + 8$. Setting $f'(x) = 0$: $3x^2 - 10x + 8 = 0$,which factors as $(3x - 4)(x - 2) = 0$. Thus,the critical points are $x = \frac{4}{3}$ and $x = 2$. Now,we find the second derivative: $f''(x) = 6x - 10$. Evaluating at $x = \frac{4}{3}$: $f''(\frac{4}{3}) = 6(\frac{4}{3}) - 10 = 8 - 10 = -2 Evaluating at $x = 2$: $f''(2) = 6(2) - 10 = 2 > 0$. The function has a local minimum at $x = 2$. The maximum value is $f(\frac{4}{3}) = (\frac{4}{3} - 1)(\frac{4}{3} - 2)^2 = (\frac{1}{3})(-\frac{2}{3})^2 = (\frac{1}{3})(\frac{4}{9}) = \frac{4}{27}$.

The value of the function $f(x) = (x - 1)(x - 2)^2$ at its maxima is

Similar Questions

Find the semi-vertical angle of a right circular cone of a given slant height,if the volume of the cone is maximum.

The greatest and least values of $f(x)=\tan ^{-1} x-\frac{1}{2} \ln x$ on $\left[\frac{1}{\sqrt{3}}, \sqrt{3}\right]$ are

The maximum value of $[x(x-1)+1]^{\frac{1}{3}}$ for $0 \leq x \leq 1$ is

The sum of the maximum and minimum values of the function $f(x) = |5x - 7| + [x^2 + 2x]$ in the interval $[\frac{5}{4}, 2]$,where $[t]$ denotes the greatest integer function $\leq t$,is:

If $x+y=60$,$x>0$,$y>0$,then the maximum value of $x y^3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module