Locate all critical points of `h(x)=(-47-6x^2+36x)^(1/2)`

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lemjay | High School Teacher | (Level 3) Senior Educator

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`h(x) = (-47-6x^2 +36x)^(1/2)`

To determine the critical points, take the derivative of h(x). Apply the power formula which is `(u^n)'= n*u^(n-1)* u' ` .

`h'(x) = 1/2 (-47 - 6x^2+36x)^(-1/2) * (-47 - 6x^2 + 36x)'`

`h'(x) =1/2 (-47 - 6x^2+36x)^(-1/2) *(-12x+36)`

`h'(x) = (-12x+36)/(2(-47-6x^2+36x)^(1/2))`

`h'(x) = (6(-2x+6))/(2(-47-6x^2+36x)^(1/2)))`

`h'(x)=(-2x+6)/(-47-6x^2+36x)^(1/2)`

Then, set h'(x) equal to zero and solve for x.

`0 = (-2x+6)/(-47-6x^2+36x)^(1/2)`

To simplify, multiply both sides by the denominator.

`(-47-6x^2+36x)^(1/2)*0 =(-2x+6)/(-47-6x^2+36x)^(1/2)*(-47-6x^2+36x)^(1/2)`

`0=-2x+6`

`2x=6`

`x=3`

Then, substitute the value fo x to h(x).

`h(x) = (-47 - 6x^2+36x)^(1/2)= (-47-6(3^2)+36(3))^(1/2)`

`h(x) = (-47 -54+108)^(1/2) = 7^(1/2)=sqrt7`

Hence, `(3,sqrt7)` is the critical point of the given function.

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