The function f(x) = x^5 - 5x^4 + 5x^3 - 10 ha | vedclass

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

Question

(B) To find the local maximum,we first find the derivative of the function $f(x) = x^5 - 5x^4 + 5x^3 - 10$.$f'(x) = 5x^4 - 20x^3 + 15x^2$.Setting $f'(

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) To find the local maximum,we first find the derivative of the function $f(x) = x^5 - 5x^4 + 5x^3 - 10$.$f'(x) = 5x^4 - 20x^3 + 15x^2$.Setting $f'(x) = 0$ for critical points:$5x^2(x^2 - 4x + 3) = 0$.$5x^2(x - 1)(x - 3) = 0$.The critical points are $x = 0, x = 1, x = 3$.Now,we find the second derivative $f''(x) = 20x^3 - 60x^2 + 30x$.Check the sign of $f''(x)$ at the critical points:For $x = 1$: $f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10$.Since $f''(1) For $x = 3$: $f''(3) = 20(27) - 60(9) + 30(3) = 540 - 540 + 90 = 90 > 0$,so it is a local minimum.For $x = 0$: $f''(0) = 0$,and checking the sign change of $f'(x)$ around $x=0$,we see $f'(x) = 5x^2(x-1)(x-3)$,which is positive on both sides of $x=0$,so it is a point of inflection.Thus,the function has a local maximum at $x = 1$.

The function $f(x) = x^5 - 5x^4 + 5x^3 - 10$ has a local maximum when $x$ is equal to:

Similar Questions

If $y = \frac{ax - b}{(x - 1)(x - 4)}$ has a turning point $P(2, -1)$,then the values of $a$ and $b$ are

The minimum value of $\frac{(5 + x)(2 + x)}{1 + x}$ for non-negative real $x$ is

The minimum value of $\left( x^2 + \frac{250}{x} \right)$ is

The maximum value of $f(x) = \frac{\log x}{x}$ $(x > 0, x \neq 1)$ is

Given $f(x) = -\frac{x^3}{3} + x^2 \sin(1.5a) - x \sin(a) \sin(2a) - 5 \sin^{-1}(a^2 - 8a + 17)$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module