# Find the critical numbers of `F(x) =x^4/(5(x-6)^2)`

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Gven

`F(x)=x^4/(5(x-6)^2)`

`` To find the critical value, we differentiate

`F'(x)=(1/5){4x^3(x-6)^2-x^4 2(x-6)}/(x-6)^4`

`=(2/5)(x^3(x-6)(2(x-6)-2x))/(x-6)^4`

`=(-12/5)(x^3(x-6))/(x-6)^4`

`=(-12/5)(x^3/(x-4)^4)`

`F'(x)=0=>x=0`

Thus F(x) will attain critical value at x=0 .

F(0)=0

Thus critical value of F(x) is 0 (zero).

The critical points of a function y = f(x) are those where f(x) has the lowest or highest possible values. The slope of the tangent drawn at points on either side of the critical point is opposite in sign.

For the function `f(x) = x^4/(5*(x - 6)^2)` , the value of f(x) is positive for both negative as well as positive values of x as the terms x^4 and (x-6)^2 have even exponents. The minimum value of x^4 is 0. This is the minimum value of the function. Drawing a graph of f(x) clearly shows that it does not take on a value lower than 0.