What are the critical values, if any, of #f(x)=x^(4/5) (x − 3)^2#?

1 Answer
Mar 11, 2016

Answer:

They are #0#, #3#, and #6/7#.

Explanation:

#f(x)=x^(4/5) (x − 3)^2#

Critical value of #f# are values in the domain of #f# at which either #f'# does not exist or #f'(x)=0#

The domain of #f# is #RR#.

So every point at which #f'# fails to exist or #f'(x)=0# is a critical value for #f#.

Differentiate using the product rule:

#f'(x) = 4/5 x^(-1/5)(x-3)^2+x^(4/5) 2(x-3)(1)# #" "# (using the chain rule at the end)

#f'(x) = (4(x-3)^2)/(5root(5)x)+(2root(5)x^4(x-3))/1#

# = (4(x-3)^2+10x(x-3))/(5root(5)x)#

# = (2(x-3)[2(x-3)+5x])/(5root(5)x)#

# = (2(x-3)(7x-6))/(5root(5)x)#

#f'(0)# does not exist, but #0# is in the domain of #f#, so #0# is a critical value for #f#.

#f'(x) = 0# at #x=3# and at #x=6/7#, both of which are in the domain of #f#.

The critical values are #0#, #3#, and #6/7#.

We have finished the problem without looking at the graph of #f#. Having done that, it can be helpful to look at the graph.
You can zoom in and out and drag the graph around using a mouse. (It will start the same every time you return to this answer.)

graph{y=x^(4/5)(x-3)^2 [-3.01, 6.857, -0.442, 4.49]}