Find the absolute maximum and absolute minimum values of f

Talamancoeb

Talamancoeb

Answered question

2021-12-18

Find the absolute maximum and absolute minimum values of f on the given interval.
f(x)=2x36x218x+9,[2,4]
absolute minimum value
absolute maximum value

Answer & Explanation

Lynne Trussell

Lynne Trussell

Beginner2021-12-19Added 32 answers

Step 1
Finding the derivative of the given function
f(x)=ddx(2x36x218x+9)
=6x212x18
To find the critical value we have,
f'(x)=0
6x212x18=0
x22x3=0
x23x+x3=0
x(x-3)+1(x-3)=0
(x+1)(x-3)=0
x=-1 or x=3
So the values between the interval [-2,-1,3,4]
Step 2
Absolute maximum and minimum
f'(-2)=-16-24+36+9=45-40=5
f'(-1)=-2-6+18+9=19
f'(3)=54-54-54+9=-45
f'(4)=128-96-72+9=-31
From the above equations, we see that the function has a maximum value at -1 and a minimum at 3.
So,
Absolute Maximum at x=-1 is 19
Absolute Minimum at x=3 is -45
Papilys3q

Papilys3q

Beginner2021-12-20Added 34 answers

Step 1
Consider the function
f(x)=2x36x218x+9, on [-2,4]
Step 2
The given function is a polynomial so it is continuous everywhere therefore is continuous on [-2,4]
Critical points: Critical points of the function are the points where the derivative of the function either zero or does not exist.
Compute the derivative of the function
f(x)=2x36x218x+9
f(x)=6x212x18
To find the critical points put f'(x)=0.
f'(x)=0
6x212x18=0
6(x22x3)=0
x22x3=0
(x-3)(x+1)=0
x=-1,3
There are two critical points x=-1 and x=3 and both lies in the interval [-2,4].
Step 3
Evaluate the function value at the critical points and end points.
f(x)=2x36x218x+9
at x=-2
f(2)=2(2)36(2)218(2)+9
f(-2)=5
at x=-1
f(1)=2(1)36(1)218(1)+9
f(-1)=19
at x=3
f(3)=2(3)36(3)218(3)+9
f(3)=-45
at x=4
f(4)=2(4)36(4)218(4)+9
f(4)=-31
Absolute maximum value and minimum value is the largest and the smallest function value respectively.
Absolute Maximum value=19 at x=-1
Absolute Minimum value=-45 at x=-3
Step 4
Answer:
Absolute Maximum value= 19
Absolute Minimum value= -45

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