Find maximum value of slope of tangent

[NIPUL JARIWALA]

I need to find the max value of slope of tangent in closed interval [0, 5].
But my answer is wrong.
f(x)= x^3 - 6x^2 + 9x +4
f'(x) = 3x^2 -12x +9
Need to find max value of f'(x)
f"(x) = 6x-12
6x-12 = 0
x = 2
f'''(x)=6>0

And it is minimum value of f'(x)

Help me to find the answer

 

[Harry_the_cat]

Ok so f'(x) is a parabola with a minimum turning point. Because you are looking at the closed interval [0, 5], calculate what is happening at the endpoints, ie when x=0 and x=5. One of these will give the largest value for f'(x).

 

[Steven G]

In a closed interval [a, b] the extreme value can occur at a point where the derivative is 0 or undefined AND within [a, b] OR the extreme value can occur at an endpoint (or endpoints).

Can you draw three diagrams for f(x) such that on the closed interval say [5, 10]
a) the max is not at an endpoint
b) the max is at x=5
c) the max is at x=10

 

[NIPUL JARIWALA]

Thank you for your reply.