How do you find the relative extrema of #f(x)=3x^5-5x^3#?

Question

Isaiah Allen · Accepted Answer

Please see the explanation below

The function is #f(x)=3x^5-5x^3# Calculate the first derivative #f'(x)=15x^4-15x^2=15x^2(x^2-1)# The critical points are when #f'(x)=0# That is, #15x^2(x^2-1)=0# The solutions to this equation are #{(x=0),(x=-1),(x=1):}# Let's build a variation chart #color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-1##color(white)(aaaa)##0##color(white)(aaaaa)##1##color(white)(aaaa)##+oo# #color(white)(aaaa)##f'(x)##color(white)(aaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+# #color(white)(aaaa)##f(x)##color(white)(aaaaa)##↗##color(white)(aaaa)##↘##color(white)(aaaa)##↘##color(white)(aaaa)##↗# When #x=-1#, there is a relative maximum at #(-1, 2)# When #x=1#, there is a relative minimum at #(1, -2)# When #x=0#, there is an inflection point at #(0, 0)# graph{3x^5-5x^3 [-10, 10, -5, 5]}

Christopher Allers · Answer

undefined To find the relative extrema of $ f(x) = 3x^5 - 5x^3 $, you first need to find its critical points by taking the derivative of the function and setting it equal to zero. Then, you can determine whether these critical points correspond to relative maxima or minima by analyzing the sign of the second derivative at those points.

1. Take the derivative of $ f(x) $ with respect to $ x $ to find $ f'(x) $.
$$ f'(x) = 15x^4 - 15x^2 $$

2. Set $ f'(x) $ equal to zero and solve for $ x $ to find the critical points.
$$ 15x^4 - 15x^2 = 0 $$
$$ 15x^2(x^2 - 1) = 0 $$
$$ x^2(x+1)(x-1) = 0 $$

So, the critical points are $ x = -1, 0, $ and $ x = 1 $.

3. Determine the sign of the second derivative $ f''(x) $ at each critical point to classify the relative extrema.
$$ f''(x) = 60x^3 - 30x $$

4. Plug the critical points into $ f''(x) $ and analyze the sign:

- $ x = -1 $: $ f''(-1) = -90 $ (negative) implies a relative maximum.
   - $ x = 0 $: $ f''(0) = 0 $ (indeterminate).
   - $ x = 1 $: $ f''(1) = 30 $ (positive) implies a relative minimum.

Therefore, the relative maximum occurs at $ x = -1 $, and the relative minimum occurs at $ x = 1 $.