The highest value among these is the absolute maximum value of f(x) on the interval [−3,1].
To find the absolute maximum value of the function f(x)=2x^3 +3x^2 +1 on the closed interval [−3,1], you need to evaluate the function at the critical points within the interval and at the endpoints of the interval.
Find the critical points:
Critical points occur where the derivative of the function is zero or undefined.
First, find the derivative of f(x):
f′ (x)=6x^2 +6x
Set f′ (x) equal to zero and solve for x
6x^2 +6x=0
Factor out 6x:
6x(x+1)=0
This gives two solutions: x=0 and x=−1.
So, the critical points are x=0 and x=−1.
Evaluate the function at the critical points and endpoints:
Evaluate
f(x) at x=−3, x=−1, and x=1.
f(−3)=2(−3)^3 +3(−3)^2 +1
f(−1)=2(−1)^3 +3(−1)^2 +1
f(1)=2(1)^3 +3(1)^2 +1
Compare the values:
Compare the values of f(x) at the critical points and endpoints to find the absolute maximum.
f(−3)
f(−1)
f(0) (critical point )f(1)
The highest value among these is the absolute maximum value of f(x) on the interval [−3,1].