Find the relative extrema, if any, of the function. Use the Second Derivative Test if applicable. (If an answer does not exist, enter DNE.) g(x) = x3 − 15x

We have the function g(x) = x^3 -15*x

First, to find extrema, we can find the zeros of the first derivative.

g'(x) = 3*x^2 -15

g'(x) = 0 = 3*x^2 - 15

x^2 = 15/3 = 5

x = √5

x = -√5

Now, watching at the second derivative we have:

g''(x) = 6*x

so when we have

g''(√5) = 6*√5 > 0 then x = √5 is a local minimum

g''(-√5) = -6*√5 < 0, then x = -√5 is a local maximum.

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.