It is given that at x=1, the function f(x) = | vedclass

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

Question

(A) Let $f(x) = x^{4}-62x^{2}+ax+9.$First,we find the derivative of the function $f(x)$ with respect to $x$:$f'(x) = \frac{d}{dx}(x^{4}-62x^{2}+ax+9)

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(A) Let $f(x) = x^{4}-62x^{2}+ax+9.$First,we find the derivative of the function $f(x)$ with respect to $x$:$f'(x) = \frac{d}{dx}(x^{4}-62x^{2}+ax+9) = 4x^{3}-124x+a.$It is given that the function $f(x)$ attains its maximum value on the interval $[0,2]$ at $x=1.$Since $x=1$ is an interior point of the interval $[0,2],$ the derivative at this point must be zero,i.e.,$f'(1) = 0.$Substituting $x=1$ into the derivative expression:$f'(1) = 4(1)^{3}-124(1)+a = 0.$$4-124+a = 0.$$-120+a = 0.$$a = 120.$Thus,the value of $a$ is $120.$

It is given that at $x=1,$ the function $f(x) = x^{4}-62x^{2}+ax+9$ attains its maximum value on the interval $[0,2].$ Find the value of $a.$

Similar Questions

The local maximum value of the function $f(x)=-(x-2)^3(x+2)^2$ is

Let $p(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$. If $p(1)=6$ and $p(3)=2$, then $p^{\prime}(0)$ is equal to

The triangle of maximum area that can be inscribed in a given circle of radius $r$ is ...... .

Find the maximum and minimum values of the function given by $g(x) = x^{3} + 1$.

The maximum and minimum values of ${x^3} - 18{x^2} + 96x$ in the interval $(0, 9)$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module