Question

Suppose that f(x)=14x−6x3. (A) Find the average of the x values of all local maxima of f. Note: If there are no local maxima, enter -1000.

Answer 1

Answer:

Maximum at $x =\frac{\sqrt{7}}{3}$

Step-by-step explanation:

Given function,

$f(x) = 14x - 6x^3$

Differentiating with respect to x,

$f'(x) = 14 - 18x^2----(1)$

For critical values :

$f'(x) = 0$

$14 - 18x^2 =0$

$14 = 18x^2$

$x^2 = \frac{14}{18}$

$x^2=\frac{7}{9}$

$x = \pm \frac{\sqrt{7}}{3}$

Now, differentiating equation (1) again with respect to x,

$f''(x) = -36x$

Since,

$f''(\frac{\sqrt{7}}{3}) = -36(\frac{\sqrt{7}}{3}) < 0$

This means that the function is maximum at $x=\frac{\sqrt{7}}{3}$

While,

$f''(-\frac{\sqrt{7}}{3}) = 36(\frac{\sqrt{7}}{3}) > 0$

This means that the function is minimum at $x=-\frac{\sqrt{7}}{3}$