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

They are

Differentiate using the product rule:

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

By signing up, you agree to our Terms of Service and Privacy Policy

To find the critical values of ( f(x) = x^{\frac{4}{5}} (x - 3)^2 ), we need to first find the derivative of ( f(x) ), set it equal to zero, and solve for ( x ).

Taking the derivative of ( f(x) ) using the product rule:

( f'(x) = \frac{4}{5}x^{-\frac{1}{5}}(x-3)^2 + x^{\frac{4}{5}}(2(x-3)) )

Setting ( f'(x) ) equal to zero and solving for ( x ):

( \frac{4}{5}x^{-\frac{1}{5}}(x-3)^2 + x^{\frac{4}{5}}(2(x-3)) = 0 )

( \frac{4}{5}x^{-\frac{1}{5}}(x-3)^2 = -2x^{\frac{4}{5}}(x-3) )

( \frac{4}{5}(x-3) = -2x^{\frac{4}{5}} )

( 4(x-3) = -\frac{10}{5}x^{\frac{4}{5}} )

( 4x - 12 = -2x^{\frac{4}{5}} )

( 2x^{\frac{4}{5}} + 4x - 12 = 0 )

This equation can be solved numerically to find the critical values.

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you find the critical points of: #f(x)=(x)(e^(-x^2))#?
- Is #f(x)=(x+1)(x+2)(x-2)# increasing or decreasing at #x=0#?
- How do you find the intervals of increasing and decreasing using the first derivative given #y=-(x^2+8x+12)#?
- What are the absolute extrema of #f(x)=x-sqrt(5x-2) in(2,5)#?
- How do you find the relative extrema for #f(x)=x^3-4x^2+x+6#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7