# `f(x) = x^2 + 3x - 8` Find all relative extrema, use the second derivative test where applicable.

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### 2 Answers

`f = x^2+3x-8`

First derivative to find the locations of extreme points,

`f' = 2*x+3-0 = 2x+3`

At extremes points, `f' = 0, 2x+3 = 0` therefore, `x = -3/2`

Now, we know that there is an extreme point at `x =-3/2,`

at `x = -3/2, f = (-3/2)^2+3*(-3/2)-8 ---gt f = 9/4-9/2-8 = (9-18-32)/4 = -41/4`

Therefore, `(-3/2, -41/4)` is a local extreme point.

Second derivative test,

`f'' = 2*1+0 = 2`

Therefore, `f"gt0` at all x. Thus` f" gt0` at `x = -3/2.`

This suggests that, the extreme point at x=-3/2 is a minimum.

Therefore, f has a local minimum of `(-3/2, -41/4)`

Find the first derivative.

Given,

`f(x)=x^2 + 3x - 8`

then,

`f'(x)=2x+3 `

Now we need to find the value(s) of x that make the first derivative zero to find the critical points.

`0=2x+3 `

`x=-3/2 ` (critical point)

This will be a relative extrema if it changes signs, so find 2 values around it to test using the first derivative test, like -2 and -1

`f'(-2)=-1 `

and ` `

`f'(-1)=1 `

Find the y-value for the x value `(-3/2,-41/4) `

Since the sign changes then the point `-3/2,-41/4 ` is a relative extrema