CVU4 - Some Review Problems: Applications of the Derivative 1. Explain why $$ f(x)=\frac{2 x-3}{x^{2}-1} $$ does not have a guraranteed extreme value on the interval $$ [-2,0] $$.

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1. The function is given by
$$
f(x)=\frac{2x-3}{x^2-1}.
$$
Notice that the denominator factors as
$$
x^2-1=(x-1)(x+1).
$$
Thus, the function is undefined when
$$
x=1 \quad \text{or} \quad x=-1.
$$

2. The interval under consideration is $$[-2,0]$$. Although $$x=1$$ is not in this interval, $$x=-1$$ is. Since $$x=-1$$ makes the denominator zero, the function is discontinuous at that point.

3. The Extreme Value Theorem guarantees that a continuous function on a closed interval has both a maximum and minimum value. However, since $$f(x)$$ is not defined (and hence not continuous) at $$x=-1$$, which is within the interval $$[-2,0]$$, the conditions of the theorem are not met.

4. Therefore, the function $$f(x)=\frac{2x-3}{x^2-1}$$ does not have a guaranteed extreme value on the interval $$[-2,0]$$ because it is not continuous on the entire interval.
The function $$ f(x) = \frac{2x-3}{x^2-1} $$ does not have a guaranteed extreme value on the interval $$[-2,0]$$ because it is undefined at $$ x = -1 $$, which is within the interval.

CVU4 - Some Review Problems: Applications of the Derivative

Explain why does not have a guraranteed extreme value on the interval .

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CVU4 - Some Review Problems: Applications of the Derivative Explain why does not have a guraranteed extreme value on the interval .