What are the local extrema, if any, of f (x) =(x^3−4 x^2-3)/(8x−4)?

Sep 30, 2017

The given function has a point of minima, but surely doesnot have a point of maxima.

Explanation:

The given function is:

$f \left(x\right) = \frac{{x}^{3} - 4 {x}^{2} - 3}{8 x - 4}$

Upon diffrentiation,

$f ' \left(x\right) = \frac{4 {x}^{3} - 3 {x}^{2} + 4 x + 6}{4 \cdot {\left(2 x - 1\right)}^{2}}$

For critical points, we have to set, f'(x) = 0.

$\implies \frac{4 {x}^{3} - 3 {x}^{2} + 4 x + 6}{4 \cdot {\left(2 x - 1\right)}^{2}} = 0$

$\implies x \approx - 0.440489$

This is the point of extrema.

To check whether the function attains a maxima or minima at this particular value, we can do the second derivative test.

$f ' ' \left(x\right) = \frac{4 {x}^{3} - 6 {x}^{2} + 3 x - 16}{2 \cdot {\left(2 x - 1\right)}^{3}}$

$f ' ' \left(- 0.44\right) > 0$

Since the second derivative is positive at that point, this implies that the function attains a point of minima at that point.