+4 This question concerns the function
f(x)=1/3x^3+x^2-3x-8
Find the exact x and y coordinates of the stationary points of f
Use the first derivate test to classify the stationary points that you found in part(a)
+118609 The first derivative gives the gradient of the tangent.
At stationary points the gradient must be 0.
So find the first derivative (differentiate it) and put that equal to 0.
Solve this equaton and you will have x value.
Sub that into the initial f(x) equation to get the y value. In this case y is called f(x).
Do as much as you can of that and then I can help more.
