An absolute extremum can only exist in an open interval of $f$. A closed interval has an infinite amount of values that approach the start and end of the intervals, so it is impossible to define an exact extremum on a closed interval.

There may not be an absolute extremum in a function if the function is unbounded.

A local extremum exists at point $x$ if $f'(x) = 0$ with odd multiplicity (i.e. the function passes $y = 0$). A value of $x$ with even multiplicity (e.g. $f'(x) = (x -2)^2$) will simply bounce off the y-axis, and $f'(x)$ does not change sign at $x$.

If the sign of $f'(x)$ at $x$ changes from negative to positive, there must exist a local minimum at $(x, f(x))$.

If the sign changes from positive to negative, there must exist a local maximum at $(x, f(x))$.

If the function has a removable discontinuity, a local maximum exists when $f(x) > \lim_{c \to x} f(c)$ and a local minimum exists when $f(x) < \lim_{c \to x} f(c)$

If the function is continuous on a closed interval, then it must have both an absolute maximum and an absolute minimum.

Show that the function $f(x) = x^2$ satisfies the hypotheses of the MVT over the interval $[-2, 4]$. Then find a value of $c$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$ on this interval.

Solution:

The interval $[a, b]$ = $[-2, 4]$

$f(x) = x^2$

$f'(x) = 2x$

$\frac{f(4) - f(-2)}{4 - (-2)} = 2$

$f'(c) = 2 = 2c$

$c = 1$

In a graph, the slope of the secant line from $(a, f(a))$ to $(b, f(b))$ will be the same as the slope of the tangent line at $(c, f(c))$

Determine whether or not the function $f(x) = |x|$ satisfies the hypotheses of the MVT over the following intervals. If yes, find the values of $c$ guaranteed by the MVT.

1. $[-2, 2]$ The interval in $[-2, 2]$ is continuous, but it's not differentiable when $x = 0$. Therefore, it does not satisfy the conditions of the MVT.
2. $[0, 4]$ The interval $[0, 4]$ is continuous, and the interval $(0, 4)$ is differentiable.$f'(x) = 1$ if $x > 0$ $\frac{f(4) - f(0)}{4 - 0} = 1$

Definition: Let the function $f$ be continuous of the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$. If $f(a) = f(b)$, then there is at least one number $c$ in $(a, b)$ such that $f'(c) = 0$.

In other words, if $f(a) = f(b)$, then there is at least one point where the average slope is zero between $[a, b]$.

Show that the function $f(x) = x^3$ satisfies the hypothesis of the MVT over the interval $[0, 1]$. Find a value of $c$.

1. Determine what intervals the function $f(x) = x^3 - 27x$ is increasing and decreasing.

   $f'(x) = 3x^2 - 27 = 0$

   $3(x + 3)(x - 3) = 0$

   $x = \{ -3, 3 \}$

   $f'(-4) > 0$

   $f'(-2) < 0$

   $f'(4) > 0$

           +        -         +
       <-------|----------|------->
              -3          3
   

   $f$ is increasing on the interval $(-\infty, -3] \cup [3, \infty)$ and decreasing on the interval $[-3, 3]$

2. $f(x) = xe^{\frac{1}{x}}$ \[ f'(x) = x \cdot e^{\frac{1}{x}} \cdot -x^{-2} + e^{\frac{1}{x}} \cdot 1 = 0 \] $f'(x) = (e^\frac{1}{x})(-x^{-1} + 1)$ $f'(x) = e^\frac{1}{x}(\frac{x - 1}{x}) = 0$ $x = 1$ , and $f'(x) = \textrm{DNE}$ if $x = 0$ $f(-1) > 0$ $f(\frac{1}{2}) < 0$ $f(2) > 0$ $f$ is increasing on $(-\infty, 0) \cup [1, \infty)$ and decreasing on $(0, 1)$

1. The graph of $f'$, the derivative of the function $f$, is shown. Which of the following statements is true about $f$? (A) $f$ is decreasing for $-1 \ge x \ge 1$ (B) $f$ is increasing for $-2 \ge x \ge 0$

• Concave up on any interval where $f''(x) > 0$
• Concave down on any interval where $f''(x) < 0$

• Point of inflection - a point where the graph of a function has a tangent line and where the concavity/sign changes (i.e. the point $(x, f(x))$ such that $f''(x) = 0$)

  • A possible POI may exist when $f''(x) = 0$, but a POI must exist when the sign changes

1. Determine the concavity of $y = 2 + \sin x$ on $[0, 2\pi]$ and any points of inflection. $y' = \cos x$ $y'' = -\sin x = 0$ $x = \{ 0, \pi, 2\pi \}$ $y'' < 0$ if $0 < x < \pi$ - concavity down at $(0, \pi)$ $y'' > 0$ is $\pi < x < 2\pi$ - concavity up at $(\pi, 2\pi)$

2. Determine the concavity of $f(x) = \frac{2x}{x^2 - 4}$

   $f''(x) = \frac{4x(x^2 + 12)}{(x^2 - 4)^3}$

   $f''(0) = 0$

   $f''(\pm 2) = \textrm{DNE}$

   $f''(-3) < 0$

   $f''(-1) > 0$

   $f''(1) < 0$

   $f''(3) > 0$

       -       +      -       +
   <-------------------------------> f''(x)
          |       |       |
          2       0      4/3
   

   Concavity down on $(-\infty, -2) \cup (0, 2)$

   Concavity up on $(-2, 0) \cup (2, \infty)$

   3 possible points of inflection: $\{ -2, 0, 2 \}$

   Since $\pm 2$ does not exist in the domain of the original function, they could not be considered points of inflection. Therefore, the only point of inflection is $(0, 0)$.

3. $f'(x) = 4x^3 - 8x^2$

   $f''(x) = 12x^2 - 16x$

   $f''(x) = 4x(3x - 4)$

   $x = \{0, \frac{4}{3}\}$

   $f''(x) > 0; x = -1$

   $f''(x) < 0; x = 1$

   $f''(x) > 0; x = 2$

         +         -          +
     <-------------------------------> f''(x)
           |         |
               0        4/3
   

   Concavity down on $(0, \frac{4}{3})$

   Convacity up on $(-\infty, 0) \cup (\frac{4}{3}, \infty)$

1. Let $f$ be the function given by $f(x) = 2xe^x$. The graph of $f$ is concave down when

   $f'(x) = 2e^x + 2xe^x$

   $f''(x) = 2e^x + 6xe^x$

   $x = \{ -2, 0 \}$

            -
     <-------------------------------> f''(x)
                |      |
               -2      0
   

   Concave down $(-\infty, -2)$

2. The second derivative of the function $f$ is given by $f''(x) = x(x - a)(x - b)^2$. The graph of $f''(x)$ is shown. For what values does $f$ have a point of inflection? $f$ has a point of inflection at $0, a, b$ because the value of $f$ at those points is zero. By definition, it is a point of inflection.
3. The function $f$ has first derivative given by $f'(x) = \frac{\sqrt{x}}{1 + x + x^3}$ $f''(x) = \frac{-5x^3 - x + 1}{2\sqrt{x}(1 + x + x^3)^2} = 0$ $-5x^3 - x + 1 = 0$ Solve the numerator for 0 $x = 0.473$ (x-coordinate of point of inflection)
4. For all $x$ in the closed interval $[2, 5]$, the function $f$ has a positive first derivative and a derivative. Which of the following could be a table of values for $f$? B is most likely the answer because $f$ in that interval is concave down. Therefore, the rate of change should slow down but not be negative.

1. Find where the extreme values of $f(x) = x^3 - 12x - 5$ occur. \begin{array}{ r l } f'(x) & = 3x^2 - 12 \\ & = 3(x^2 - 4) \\ & = 3(x + 2)(x - 2) \\ \end{array} $x = \pm 2$ $f''(x) = 6x$ $f''(2) = 12$ — a local maximum occurs at $x = 2$ $f''(-2) = -12$ — a local minumum occurs at $x = -2$

1. Let $g$ be a twice-differentiable function with $g' > 0$ and $g''(x) > 0$ for all real numbers $x$, such that $g(4) = 12$ and $g(5) = 18$. Which is a possible value of $g(6)$?

   1. 15
   2. 21
   3. 24
   4. 27

   The most likely answer is (D) because $g'$ is positive, and $g''$ is also positive, which means that $g$ must be exponentially increasing.

2. The function $f$ has property $f(x), f'(x), f''(x)$ are negative for all real values of $x$. Which of the following could be the graph for $f$?

Graph (B) is most likely the answer. The range of $f$ is always negative, so it can never be positive at any point. Since $f'$ is negative, $f$ should be decreasing. Since $f''$ is negative, $f'$ should be decreasing, which means $f$ should be exponentially decreasing (i.e. no deceleration).

Sketch the graph of $f(x) = \frac{2x}{x^2 - 4}$

x-intercept: $(0, 0)$

y-intercept: $(0, 0)$

Vertical asymptote: $x = \pm 2$

Horizontal asymptote: $\lim_{x \to \pm \infty} f(x) = 0$

$f(x) = 2x(x^2 - 4)^{-1}$

$f'(x) = \frac{-2x(x^2 + 4)}{(x^2 - 4)^2}$

$f''(x) = \frac{4x(x^2 + 12)}{(x^2 - 4)^3}$

$f'(0) = 0$

$f'(\pm 2) = \textrm{undefined}$

$f''(0) = 0$

$f''(\pm2) = \textrm{undefined}$

        -          -         -         -
 <-------------------------------------------> f'(x)
              2         0         2

       -          +         -         +
<-------------------------------------------> f''(x)
             2         0         2

                  Decreasing
     CCU         CCD       CCU       CCD
<-------------------------------------------> f(x)
             2         0         2

A particle moves along a horizontal axis with velocity given by the function

\[ v(t) = -5 + 2.3e^{\sin(t^2)} \]

Is the particle speeding up or slowing down at $t = 27$?

\[ v'(t) = 2.3(e^{\sin(t^2)} + -2t\cos(t^2)) \]

\[ a(27) = v'(27) = -120 \]

Because $a(27) < 0$, the particle is decelerating, and which means the particle is slowing down.