What are the local extrema of #f(x)= (x^3-x^2-5x+4)/(x-2)^2#?

Question

Scarlett Allison · Accepted Answer

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Scarlett Allison · Answer

undefined To find the local extrema of $ f(x) = \frac{x^3 - x^2 - 5x + 4}{(x - 2)^2} $, we need to first find its derivative and then solve for critical points where the derivative is zero or undefined. Then, we can determine if these points correspond to local extrema by using the first or second derivative test.

First, find the derivative of $ f(x) $ using the quotient rule:

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

After simplifying, we get:

$$ f'(x) = \frac{3x^4 - 10x^3 + 9x^2 - 8x - 8}{(x - 2)^3} $$

To find critical points, set $ f'(x) $ equal to zero and solve for $ x $.

Solving $ f'(x) = 0 $ yields $ x = 2 $ and $ x \approx -0.588 $.

Now, we can apply the first or second derivative test to determine if these critical points correspond to local extrema.

We find that $ f''(x) = \frac{12x^3 - 36x^2 + 54x - 32}{(x - 2)^4} $.

At $ x = 2 $, $ f''(2) = \frac{3}{2} $, which is positive, indicating a local minimum.

At $ x \approx -0.588 $, $ f''(-0.588) \approx -0.79 $, which is negative, indicating a local maximum.

Therefore, the local extrema of $ f(x) $ are:

- Local minimum at $ x = 2 $
- Local maximum at $ x \approx -0.588 $