§3.3 Using derivatives to identify extreme values

Introduction

Critical numbers and the first derivative test

The second derivative test

A graphed function $h$ falls from $h(-3) = 3$ to a corner low point, rises to a peak at $(0, 2)$, falls to a smooth minimum, and climbs to $h(3) = 1.25$. On the open interval $-3 \lt c \lt 3$, where does $h$ have a local maximum?

Let $f'(x) = e^{-2x}(3-x)(x+1)^2$. The critical numbers of $f$ and their classifications are

Suppose $g'(x) = \frac{(x+4)(x-2)}{x^2+1}$. The critical numbers of $g$ are

For that same $g$, the sign chart of $g'$ reads positive on $x \lt -4$, negative on $-4 \lt x \lt 2$, and positive on $x \gt 2$. At $x = -4$ the function $g$ has

With the same sign chart, which full statement about the extrema of $g$ is correct?

For the same $g$, note that $\lim_{x \to \pm\infty} g'(x) = 1$. A sketch of $y = g(x)$ must therefore

Let $f'(x) = 3x^4 - 9x^2 = 3x^2(x-\sqrt{3})(x+\sqrt{3})$. The sign chart of $f'$ shows

A graph of $g''$ shows $g'' \gt 0$ for $x \lt -1$, $g'' \lt 0$ for $-1 \lt x \lt 2$, and $g'' \lt 0$ for $x \gt 2$. The function $g$ is concave up on

For $g'(x) = \frac{(x+4)(x-2)}{x^2+1}$, compute $g'(2)$.

The graph of $g''$ shows $g'' \gt 0$ for $x \lt -1$ and $g'' \lt 0$ for $x \gt -1$. State the $x$-coordinate of the inflection point of $g$.

Suppose $g'(-1.67857351) = 0$ and the graph of $g''$ shows $g''(-1.67857351) \gt 0$. Then at that input $g$ has

The graph of $g''$ shows a simple zero at $x = -1$ and a repeated zero at $x = 2$, and $g''$ is assumed to be a polynomial. The degree of $g$ is likely

For $h(x) = x^2 + \cos(kx)$, plotting several $k$ values shows visible concavity changes appearing near $x \approx \pm 0.5$ for $k = 1.6$ but nothing detectable for smaller $k$. The smallest $k$ at which inflection points become visible is about

For $h(x) = x^2 + \cos(kx)$, the second derivative is $h''(x) = 2 - k^2\cos(kx)$. The graph has no inflection points for $k \le \sqrt{2}$ and infinitely many for $k \gt \sqrt{2}$ because

For the same family, critical numbers solve $2x = k\sin(kx)$. No matter the value of $k$, there are only finitely many because

A critical number of $f$ where $f'$ does not change sign can fail to be the location of any local maximum or local minimum.