SOLUTION: 𝑓(𝑥) = 3𝑥^5 − 5𝑥^3 defined on [-3, 3] Identify the optimal point(s) of this function. Find the global max and min points of this function.

Algebra ->  Functions -> SOLUTION: 𝑓(𝑥) = 3𝑥^5 − 5𝑥^3 defined on [-3, 3] Identify the optimal point(s) of this function. Find the global max and min points of this function.       Log On


   

Question 1170622: 𝑓(𝑥) = 3𝑥^5 − 5𝑥^3 defined on [-3, 3]
Identify the optimal point(s) of this function.
Find the global max and min points of this function.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+3x%5E5+-+5x%5E3 defined on [-3, 3]
Identify the optimal point(s) of this function.
Find the global max and min points of this function.
x-intercepts:
0+=+%283x%5E2+-+5%29x%5E3-> x=0 or
0+=+%283x%5E2+-+5%29 ->5+=+%283x%5E2+-+5%29 ->5%2F3+=+x%5E2 -> x=%2B-sqrt%285%2F3%29+
x-intercepts are at: (sqrt%285%2F3+%29,0) and (-sqrt%285%2F3+%29,0)
y-intercept:
f%28x%29+=+3%2A0%5E5+-+5%2A0%5E3=0
y-intercept at: (0,0)
global max :
use first derivative test

%28d%2Fdx%29%283x%5E5+-+5x%5E3%29+=5%2A3x%5E4-3%2A5x%5E2
%28d%2Fdx%29%283x%5E5+-+5x%5E3%29+=15x%5E4-15x%5E2
%28d%2Fdx%29%283x%5E5+-+5x%5E3%29+=+15x%5E2%28x%5E2+-+1%29
->%28x%5E2+-+1%29=0 ->x%5E2+=+1->x=1 or x=-1

then
f%28x%29+=+3%2A1%5E5+-+5%2A1%5E3=-2
f%28x%29+=+3%2A%28-1%29%5E5+-+5%2A%28-1%29%5E3=-3-%28-5%29=-3%2B5=2


max{ +3x%5E5+-+5x%5E3 } =+2 atx+=+-1
and
min { 3x%5E5+-+5x%5E3 } =+-2 at x+=+1
use given interval as critical points:
f%28-3%29+=+3%2A%28-3%29%5E5+-+5%2A%28-3%29%5E3=-594
min (-3,+-594 )
f%283%29+=+3%2A%283%29%5E5+-+5%2A%283%29%5E3=594
max(3,594)