How do you find all critical numbers of the function #f(x)=x^(4/5)(x-4)^2#?

Question

41748 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-27T23:41:17

#x=4, x=8/7,x=0#

Critical points are points in the domain where the derivative is equal to zero or where the derivative is not defined.

Derivative Equal to Zero:

The derivative can be found by using the power rule and the chain rule.

#f'(x) = \frac{14x^2-72x+64}{5x^{\frac{1}{5}}}#

Now we set this equal to zero:

#\frac{14x^2-72x+64}{5x^{\frac{1}{5}}}=0#

#14x^2-72x+64=0#

#2(7x^2-36x+32)=0#

#2(7x-8)(x-4)=0#

We get:

#7x-8=0rarrx=8/7#
#x-4=0rarrx=4#

Derivative Not Defined:

The only possible case where # \frac{14x^2-72x+64}{5x^{\frac{1}{5}}}# is not defined is when the denominator is equal to zero.

Therefore we set #5x^(1/5)=0# and solve.

#5x^(1/5)=0#

#x=0#

Therefore, the derivative is not defined at #x=0#.

#0# is in the domain of #f# so #0# is a critical point for #f#.

Conclusion

The critical points of this function are #x=4, x=8/7,x=0#