Suppose that $\displaystyle{f'(x) = \frac{-1}{x^2+1}}$. If $f(a) \lt f(b)$, which is greater, $a$ or $b$? Justify your answer.

Note that $f'(x) \lt 0$ for all values of $x$. The mean value theorem tells us that $(b-a)f'(c) = f(b)-f(a)$ for some value $c$. If $f(a) \lt f(b)$, then $f(b)-f(a)$ is positive. For the equation to make sense, we then must have $b-a$ negative. Thus, $a \ge b$.

Suppose that $f'(x) = 2^x$. If $a \lt b$, which is greater, $f(a)$ or $f(b)$? Justify your answer.

Note $f'(x)$ is always positive, and $a \lt b$ means that $b-a$ is positive. Thus, as the mean value theorem tells us that $(b-a)f'(c) = f(b)-f(a)$ for some value $c$, it must be the case that $f(b) - f(a)$ is positive. Consequently, $f(b) \gt f(a)$.

Determine whether the Mean Value Theorem (MVT) can be applied to the given functions and interval. If not, justify what condition from the MVT fails.

$\displaystyle{f(x) = \frac{x+4}{x-1}; \quad [2,5]}$

$\displaystyle{f(x) = \frac{x+4}{x-1}; \quad [0,5]}$

$\displaystyle{f(x) = |1-x|; \quad [-1,1]}$

$\displaystyle{1-|x|; \quad [-1,1]}$

$\displaystyle{\frac{x^{1/3}}{x+2}; \quad [-1,8]}$

MVT applies

MVT does not apply; there is an infinite discontinuity at $x=1$

MVT applies

MVT does not apply; there is a corner at $x=0$ (so $f$ is not differentiable there)

MVT does not apply; the function is continuous but its simplified derivative is $\displaystyle{f'(x) = \frac{-2(x+1)}{3x^{2/3}(x+2)^2}}$, so the function has a vertical tangent at $x=0$

Suppose that $f(-2)=4$ and $f'(x) \lt -2$ for all $x \gt -2$. Will $f(0)$ be a positive or negative number?

Using the Mean Value Theorem, we conclude that $\displaystyle{f'(c) = \frac{f(-2)-f(0)}{-2-0} = \frac{4 - f(0)}{-2}}$.

Since $f'(x) \lt -2$ for all $x$, this means that $f'(c) \lt -2$ as well.

So $\displaystyle{\frac{4-f(0)}{-2} \lt -2}$, which means $4-f(0) \gt 4$ (remember to change the inequality sign when you multiply by a negative number!).

Hence, $f(0) \lt 0$.