What are the local extrema of #f(x)= x^3-7x#?

Question

Ezekiel Alexander · Accepted Answer

Turning points (local extrema) occur when the derivative of the function is zero,
ie when #f'(x)=0#.
that is when #3x^2-7=0#
#=>x=+-sqrt(7/3)#. since the second derivative #f''(x)=6x#, and 
#f''(sqrt(7/3))>0 and f''(-sqrt(7/3))<0#, it implies that #sqrt(7/3) # is a relative minimum and #-sqrt(7/3)# is a relative maximum. The corresponding y values may be found by substituting back into the original equation. The graph of the function makes verifies the above calculations. graph{x^3-7x [-16.01, 16.02, -8.01, 8]}

Axel Alvarez · Answer

undefined To find the local extrema of $ f(x) = x^3 - 7x $, we first find its derivative, $ f'(x) $, and then solve for points where $ f'(x) = 0 $ or where $ f'(x) $ is undefined.

$$ f'(x) = 3x^2 - 7 $$

Setting $ f'(x) $ equal to zero and solving for $ x $, we get:

$$ 3x^2 - 7 = 0 $$
$$ 3x^2 = 7 $$
$$ x^2 = \frac{7}{3} $$
$$ x = \pm \sqrt{\frac{7}{3}} $$

The critical points are $ x = -\sqrt{\frac{7}{3}} $ and $ x = \sqrt{\frac{7}{3}} $.

To determine the nature of the extrema, we examine the sign of the derivative around these critical points.

When $ x < -\sqrt{\frac{7}{3}} $, $ f'(x) $ is positive, indicating $ f(x) $ is increasing.
When $ -\sqrt{\frac{7}{3}} < x < \sqrt{\frac{7}{3}} $, $ f'(x) $ is negative, indicating $ f(x) $ is decreasing.
When $ x > \sqrt{\frac{7}{3}} $, $ f'(x) $ is positive again, indicating $ f(x) $ is increasing.

So, we have a local maximum at $ x = -\sqrt{\frac{7}{3}} $ and a local minimum at $ x = \sqrt{\frac{7}{3}} $.