Let h(x) = frac{5(f(x))^3}{3} + frac{(f(x))^2 | vedclass

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

Question

(C) Given $h(x) = \frac{5}{3}(f(x))^3 + \frac{1}{2}(f(x))^2 + 2f(x) + 100$. To determine the behavior of $h(x)$,we differentiate with respect to $x$:

Vedclass · Accepted Answer

$h(x)$ increases as $f(x)$ increases

Vedclass · Answer

(C) Given $h(x) = \frac{5}{3}(f(x))^3 + \frac{1}{2}(f(x))^2 + 2f(x) + 100$. To determine the behavior of $h(x)$,we differentiate with respect to $x$: $h'(x) = \frac{d}{dx} \left[ \frac{5}{3}(f(x))^3 + \frac{1}{2}(f(x))^2 + 2f(x) + 100 \right]$. Using the chain rule: $h'(x) = \left( 5(f(x))^2 + f(x) + 2 \right) f'(x)$. Consider the quadratic expression $Q = 5(f(x))^2 + f(x) + 2$. The discriminant $D = b^2 - 4ac = (1)^2 - 4(5)(2) = 1 - 40 = -39$. Since $D  0)$,the quadratic $Q$ is always positive for all real values of $f(x)$. Therefore,the sign of $h'(x)$ depends solely on the sign of $f'(x)$. If $f'(x) > 0$,then $h'(x) > 0$,meaning $h(x)$ increases as $f(x)$ increases. Thus,$h(x)$ increases as $f(x)$ increases.

Let $h(x) = \frac{5(f(x))^3}{3} + \frac{(f(x))^2}{2} + 2f(x) + 100$,where $f(x)$ is a differentiable function. Then which one of the following is correct?

Similar Questions

If $y = \sin (\sqrt {\sin x + \cos x} )$,then $\frac{dy}{dx} = $

If $y = \frac{x}{2}\sqrt{a^2 + x^2} + \frac{a^2}{2}\log(x + \sqrt{x^2 + a^2})$,then $\frac{dy}{dx} = $

If $f(x) = (x - x_0)g(x)$,where $g(x)$ is continuous at $x_0$,then $f'(x_0)$ is equal to

Find the value of $k$ if $\frac{d}{d x}\left\{\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos (4 x)}}}}\right\} = k \sec \left(\frac{x}{2}\right) \tan \left(\frac{x}{2}\right)$

If $f(x)$ is a polynomial such that $f(x) = [f'(x)]^2$ and $f(2) = 0$, then $f(-2) = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module