`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.
`f(x)=x^2 + 3x - 8`
Now we need to find the value(s) of x that make the first derivative zero to find the critical points.
`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
and ` `
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