For each of the following functions, find the maximum and...

For each of the following functions, find the maximum and minimum values on the indicated intervals, by finding the points in the interva? where the derivative is $0,$ and comparing the values at these points with the values at the end points. (i) $\quad f(x)=x^{3}-x^{2}-8 x+1$ on [-2,2] (ii) $f(x)=x^{5}+x+1$ on [-1,1] (iii) $f(x)=3 x^{4}-8 x^{3}+6 x^{2} \quad$ on $\left[-\frac{1}{2}, \frac{1}{2}\right]$ (iv) $f(x)=\frac{1}{x^{5}+x+1}$ on $\left[-\frac{1}{2}, 1\right]$ (v) $f(x)=\frac{x+1}{x^{2}+1}$ on $\left[-1, \frac{1}{2}\right]$ (vi) $f(x)=\frac{x}{x^{2}-1}$ on [0,5]

For each of the following functions, find the maximum and minimum values on the indicated intervals, by finding the points in the interva? where the derivative is $0,$ and comparing the values at these points with the values at the end points.
(i) $\quad f(x)=x^{3}-x^{2}-8 x+1$ on [-2,2]
(ii) $f(x)=x^{5}+x+1$ on [-1,1]
(iii) $f(x)=3 x^{4}-8 x^{3}+6 x^{2} \quad$ on $\left[-\frac{1}{2}, \frac{1}{2}\right]$
(iv) $f(x)=\frac{1}{x^{5}+x+1}$
on $\left[-\frac{1}{2}, 1\right]$
(v) $f(x)=\frac{x+1}{x^{2}+1}$ on $\left[-1, \frac{1}{2}\right]$
(vi) $f(x)=\frac{x}{x^{2}-1}$ on [0,5]

Key Concepts

Extreme Value Theorem

The Extreme Value Theorem states that if a function is continuous on a closed interval, then it must attain both a maximum and a minimum value on that interval. This theorem forms the foundation for many optimization problems in calculus, ensuring that solutions actually exist when you search within a closed domain.

Critical Points

Critical points are points in the domain of a function where the derivative is zero or undefined. These points are important because they are potential candidates for local extrema. Identifying critical points is a key step in finding the maximum and minimum values of a continuous function on a closed interval.

Derivative Analysis in Optimization

Taking the derivative of a function is essential in optimization problems to find where the rate of change is zero, indicating potential peaks or valleys. By setting the derivative equal to zero and solving, you find critical points that, along with the interval endpoints, must be examined to determine the function's extreme values.

Endpoint Evaluation

In addition to evaluating the function at the critical points, it is necessary to evaluate it at the endpoints of the closed interval. This is because even if the derivative does not vanish at the endpoints, these points might still yield the highest or lowest function values, as implied by the Extreme Value Theorem.

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