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

1 Answer
Jim H
Mar 11, 2016

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]}

Related questions
See all questions in Identifying Stationary Points (Critical Points) for a Function
Impact of this question
7157 views around the world
You can reuse this answer
Creative Commons License