For the function f(x) = x^{40} - x^{20},find | vedclass

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

Question

(B) To find the absolute minimum of $f(x) = x^{40} - x^{20}$ on the interval $[0, 1]$,we first find the critical points by setting the derivative $f'(

Vedclass · Accepted Answer

$-1/4$

Vedclass · Answer

(B) To find the absolute minimum of $f(x) = x^{40} - x^{20}$ on the interval $[0, 1]$,we first find the critical points by setting the derivative $f'(x) = 0$.$f'(x) = 40x^{39} - 20x^{19} = 20x^{19}(2x^{20} - 1)$.Setting $f'(x) = 0$,we get $x = 0$ or $2x^{20} = 1$,which implies $x^{20} = 1/2$,so $x = (1/2)^{1/20}$.Since $(1/2)^{1/20}$ is in the interval $(0, 1)$,we evaluate $f(x)$ at the critical point and the endpoints $x = 0$ and $x = 1$.$f(0) = 0^{40} - 0^{20} = 0$.$f(1) = 1^{40} - 1^{20} = 0$.$f((1/2)^{1/20}) = ((1/2)^{1/20})^{40} - ((1/2)^{1/20})^{20} = (1/2)^2 - (1/2)^1 = 1/4 - 1/2 = -1/4$.Comparing the values $0, 0,$ and $-1/4$,the absolute minimum value is $-1/4$.

For the function $f(x) = x^{40} - x^{20}$,find the absolute minimum value in the interval $[0, 1]$.

Similar Questions

The sum of two non-zero numbers is $4$. The minimum value of the sum of their reciprocals is

The function $f(x) = \sin x(1 + \cos x)$ at $x = \frac{\pi}{3}$ is:

Find all the points of local maxima and local minima of the function $f$ given by $f(x) = 2x^3 - 6x^2 + 6x + 5$.

$A$ metal wire $108 \ m$ long is bent to form a rectangle. If the area of the rectangle is maximum,then its dimensions are

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module