`f(x) = ln(x^2 + x + 1), [-1, 1]` Find the absolute maximum and minimum values of f on the given interval

Textbook Question

Chapter 4, 4.1 - Problem 61 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

The function is defined because x^2+x+1 always >0.

Further, ln(y) is monotone increasing and therefore it has min and max where y has min and max.

x^2+x+1 has a global minimum at x=-1/2 and it is 3/4, and global maximum at x=1 (it is 3). So f has minimum ln(3/4) and maximum ln(3) . 

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scisser | (Level 3) Honors

Posted on

Take the derivative:

`f'(x)=(2x+1)/(x^2+x+1) `

Set the numerator equal to 0.

`2x+1=0 `

`2x=-1 `

`x=-1/2 `

Find the boundaries

`f(-1)=0 `

`f(-1/2)=ln0.75 `

`f(1)=ln3 `

Therefore,

the absolute max is `ln3 ` at `x=1 `

the absolute min is `ln0.75 ` at `x=-1/2 `

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