`f(x) = x^3 - 6x^2 + 9x + 1, [2,4]` Find the local and absolute extreme values of the function on the given interval.

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Chapter 4, Review - Problem 1 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`f(x)=x^3-6x^2+9x+1,[2,4]`

Now to find the absolute extreme values of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.

`f'(x)=3x^2-6(2x)+9`

`f'(x)=3x^2-12x+9`

Now to find the critical numbers, solve for x for f'(x)=0.

`3x^2-12x+9=0`

`3(x^2-4x+3)=0`

`x^2-4x+3=0`

`x^2-3x-x+3=0`

`x(x-3)-1(x-3)=0`

`(x-3)(x-1)=0`

`x=3, x=1`

However x=1 is not in the domain of the function, so the critical point is at x=3,

Now let's evaluate the function at the end points and at the critical number,

`f(2)=2^3-6(2^2)+9*2+1=3`

`f(4)=4^3-6(4^2)+9*4+1=5`

`f(3)=3^3-6(3^2)+9*3+1=1`

Absolute Maximum=5 at x=4

Absolute Minimum=1 at x=3

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