The minimum value of the function f(x) = frac | vedclass

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

Question

(D) Let $g(x) = 3x^4 + 8x^3 - 18x^2 + 60$. The function $f(x) = \frac{40}{g(x)}$ is minimum when $g(x)$ is maximum,and $f(x)$ is maximum when $g(x)$ i

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let $g(x) = 3x^4 + 8x^3 - 18x^2 + 60$. The function $f(x) = \frac{40}{g(x)}$ is minimum when $g(x)$ is maximum,and $f(x)$ is maximum when $g(x)$ is minimum.To find the extrema of $g(x)$,we find its derivative: $g'(x) = 12x^3 + 24x^2 - 36x$.Setting $g'(x) = 0$,we get $12x(x^2 + 2x - 3) = 0$,which factors to $12x(x+3)(x-1) = 0$.The critical points are $x = 0, x = 1, x = -3$.We evaluate $g(x)$ at these points:$g(0) = 60$.$g(1) = 3(1)^4 + 8(1)^3 - 18(1)^2 + 60 = 3 + 8 - 18 + 60 = 53$.$g(-3) = 3(-3)^4 + 8(-3)^3 - 18(-3)^2 + 60 = 3(81) + 8(-27) - 18(9) + 60 = 243 - 216 - 162 + 60 = -75$.Since $g(x) 	o \infty$ as $x 	o \pm \infty$,the global minimum of $g(x)$ is $-75$ at $x = -3$. However,$f(x)$ is defined as $40/g(x)$.Checking the values of $f(x)$ at critical points:$f(0) = 40/60 = 2/3$.$f(1) = 40/53$.$f(-3) = 40/(-75) = -8/15$.As $x 	o \pm \infty$,$f(x) 	o 0$.The minimum value of the function is $-8/15$. Since this is not among the options,the correct choice is $D$.

The minimum value of the function $f(x) = \frac{40}{3x^4 + 8x^3 - 18x^2 + 60}$ is:

Similar Questions

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

$x$ and $y$ are two variables such that $x > 0$ and $xy = 1$. Then the minimum value of $x + y$ is

Find the point where $f(x) = 2\sin x + \cos 2x$ has a maximum value in the interval $0 \leq x \leq 2\pi$.

The function $f(x) = \frac{x}{1 + x \tan x}$ attains its maximum value at:

If $y=a \log x+b x^2+x$ has its extremum values at $x=-1$ and $x=2$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module