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

The sum of the terms of an infinitely decreasing geometric progression is equal to the greatest value of the function $f(x) = x^3 + 3x - 9$ on the interval $[-2, 3]$. If the difference between the first and the second term of the progression is equal to $f'(0)$,then the common ratio of the $G.P.$ is

The number that exceeds its square by the greatest amount is

The ordinate of all points on the curve $y = \frac{1}{2\sin^2 x + 3\cos^2 x}$ where the tangent is horizontal,is

The number of distinct real roots of the equation $3x^{4} + 4x^{3} - 12x^{2} + 4 = 0$ is ..... .

$A$ wire of length $20 \text{ cm}$ is bent in the form of a sector of a circle. The maximum area that can be enclosed by the wire is (in $\text{ cm}^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module