Question

i asked about stationary points which came out at (x-3) and (x+1)

i am now trying to use the first derivative test to classify the left hand point...

(x-3)
so picking an xL point of -4 and an xR point of -1

i get this
f ' (-4) = 3 16 - 4 + 9 = 48 - 24 + 9 = 33 >0
f ' (-1) = 3 1 - 6 1 + 9 = 3 - 6 + 9 = 6 >0

but this means it isn't a local max or min so i am confused i know my calculation is wrong somewhere...

Answer 1

f(x)=-x^3+3x^2+9x-5 asked to find the stationary points

f'=-3x^2+6x+9=0

leads to (x-3)(x+1)=0
or x=3 or -1

so at x=3, try x=2 and x=4

f'(4)=(4-3)(4+1)=5>0
f'(2)=(2-3)(2-1)=-1<0

Now for the x=-1 point, choose x=0 and -2

f'(0)=(-3)(+1)=-3
f'(-2)=(-2-4)(-2+1)=+8