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

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Answer 1 · 2017-09-30T08:31:07

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

The given function is:

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

Upon diffrentiation,

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

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

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

$\Rightarrow x \approx - 0.440489$ $implies x ~~ -0.440489$

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.

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

$f'' (- 0.44) > 0$ $f''(-0.44) > 0$

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

What are the local extrema, if any, of f(x)=x3−4x2−38x−4f (x) =(x^3−4 x^2-3)/(8x−4)?