The function is f(x)=2x3−15x2+36x+1 on the interval [1,5].
To find the absolute minimum value, we need to check the critical points and the endpoints of the interval.
Find the derivative:
f′(x)=6x2−30x+36
Set equal to zero:
6x2−30x+36=0
x2−5x+6=0
(x−2)(x−3)=0
The critical points are x=2 and x=3. Both lie within [1,5].
Evaluate f(x) at x=1:
f(1)=2(1)3−15(1)2+36(1)+1
f(1)=2−15+36+1
f(1)=24
Evaluate f(x) at x=2:
f(2)=2(2)3−15(2)2+36(2)+1
f(2)=16−60+72+1
f(2)=29
Evaluate f(x) at x=3:
f(3)=2(3)3−15(3)2+36(3)+1
f(3)=54−135+108+1
f(3)=28
Evaluate f(x) at x=5:
f(5)=2(5)3−15(5)2+36(5)+1
f(5)=250−375+180+1
f(5)=56
Comparing all values: f(1)=24, f(2)=29, f(3)=28, f(5)=56
The absolute minimum value of f(x) on [1,5] is 24.