Graphing Techniques

Recall that given a function $f(x)$ , we defined its second derivative by the first derivative of the first derivative:

$f''(x) := (f')'(x).$

In Leibniz notation,

$\displaystyle \frac{\mathrm d^2 f}{\mathrm dx^2}(x) \equiv \frac{\mathrm d^2}{\mathrm dx^2} (f(x)) := \frac{\mathrm d}{\mathrm dx}\left( \frac{\mathrm d}{\mathrm dx}(f(x)) \right).$

Question 1. Given constants $m, c$ , evaluate

$\displaystyle \frac{\mathrm d^2}{\mathrm dx^2} (mx+c).$

(Click for Solution)

Solution. Differentiating twice,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} (mx+c) &= m \cdot \frac{\mathrm d}{\mathrm dx}(x) + \frac{\mathrm d}{\mathrm dx}(c) \\ &= m \cdot 1 + 0 \\ &= m, \\ \frac{\mathrm d^2}{\mathrm dx^2} (mx+c) &= \frac{\mathrm d}{\mathrm dx} \left( \frac{\mathrm d}{\mathrm dx} (mx+c)\right) \\ &= \frac{\mathrm d}{\mathrm dx} (m) \\ &= 0. \end{aligned}$

Question 2. Consider the graphs of $y = f(x)$ and $y = g(x)$ below.

Explain qualitatively why $f''(a) > 0$ and $g''(b) < 0$ . In this case, we say that $f$ is convex near $a$ and $g$ is concave near $b$ .

(Click for Solution)

Solution. Let $T_c$ denote the tangent to $y = f(x)$ at $(c, f(c))$ with gradient $f'(c)$ . For some small $\delta > 0$ , $T_{a+\delta}$ is steeper than $T_{a-\delta}$ , so that $f'(a-\delta) < f'(a+\delta)$ . Since $f'$ is increasing from $a - \delta$ to $a + \delta$ , $f''(a) > 0$ .

Similarly, let $S_c$ denote the tangent to $y = g(x)$ at $(c, g(c))$ with gradient $g'(c)$ . For some small $\delta > 0$ , $S_{b-\delta}$ is steeper than $S_{b+\delta}$ , so that $g'(b-\delta) > g'(n+\delta)$ . Since $g'$ is decreasing from $b - \delta$ to $b + \delta$ , $g''(b) < 0$ .

Question 3. Given any rational number $n$ , evaluate $\displaystyle \frac{\mathrm d^2}{\mathrm dx^2}(x^n)$ .

(Click for Solution)

Solution. Differentiating twice,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} (x^n) &= n x^{n-1}, \\ \frac{\mathrm d^2}{\mathrm dx^2} (x^n) &= \frac{\mathrm d}{\mathrm dx} \left( \frac{\mathrm d}{\mathrm dx} (x^n)\right) \\ &= \frac{\mathrm d}{\mathrm dx} (nx^{n-1}) \\ &= n \cdot \frac{\mathrm d}{\mathrm dx} (x^{n-1}) \\ &= n (n-1) x^{n-2}. \end{aligned}$

Question 4. On the same diagram, sketch the graphs of $y = x^2$ and $y = x^{1/2}$ for $x \geq 0$ . What do you notice?

(Click for Solution)

Solution. Using Question 3,

$\displaystyle \frac{\mathrm d^2}{\mathrm dx^2}(x^2) = 2,\quad \frac{\mathrm d^2}{\mathrm dx^2}(x^{1/2}) = -\frac 14 \cdot x^{-3/2}.$

For $x > 0$ ,

$\displaystyle \frac{\mathrm d^2}{\mathrm dx^2}(x^2) > 0,\quad \frac{\mathrm d^2}{\mathrm dx^2}(x^{1/2}) < 0.$

Therefore, we sketch as follows.

Both graphs pass through $(0,0)$ and $(1,1)$ , and are reflections of each other about the line $y = x$ .

Question 5. On the same diagram, sketch the graphs of $y = x^3$ and $y = x^{1/3}$ for $x \neq 0$ . What do you notice?

(Click for Solution)

Solution. Using Question 3,

$\displaystyle \frac{\mathrm d^2}{\mathrm dx^2}(x^3) = 6x,\quad \frac{\mathrm d^2}{\mathrm dx^2}(x^{1/3}) = -\frac 29 \cdot x^{-5/3}.$

For $x > 0$ ,

$\displaystyle \frac{\mathrm d^2}{\mathrm dx^2}(x^3) > 0,\quad \frac{\mathrm d^2}{\mathrm dx^2}(x^{1/3}) < 0.$

For $x < 0$ ,

$\displaystyle \frac{\mathrm d^2}{\mathrm dx^2}(x^3) < 0,\quad \frac{\mathrm d^2}{\mathrm dx^2}(x^{1/3}) > 0.$

Therefore, we sketch as follows.

Both graphs pass through the points $(-1,-1)$ , $(0,0)$ , and $(1,1)$ , and are reflections of each other about the line $y = x$ .

Question 6. Given that $f(g(x)) = x$ , show that if $f'(g(a)) > 0$ , then $g'(a) > 0$ . Furthermore, if $f$ is convex near $g(a)$ , then show that $g$ is concave near $a$ .

(Click for Solution)

Solution. By the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(x) &= \frac{\mathrm d}{\mathrm dx}(f(g(x))) \\ 1 &= f'(g'(x)) \cdot g'(x). \end{aligned}$

Setting $x = a$ ,

$\displaystyle 1 = f'(g(a)) \cdot g'(a) \quad \Rightarrow \quad g'(a) = \frac 1{f'( g(a) )}.$

Therefore, $g'(a) > 0$ if $f'(g(a)) > 0$ .

By the product rule and the chain rule again,

$\begin{aligned} \frac{\mathrm d^2}{\mathrm dx^2}(x) &= \frac{\mathrm d^2}{\mathrm dx^2}(f(g(x))) \\ \frac{\mathrm d}{\mathrm dx}(1) &= \frac{\mathrm d}{\mathrm dx}(f'(g(x)) \cdot g'(x)), \\ 0 &= \frac{\mathrm d}{\mathrm dx}( f'(g(x))) \cdot g'(x) + f'(g(x)) \cdot g''(x) \\ 0 &= f''(g(x)) \cdot (g'(x))^2 + f'(g(x)) \cdot g''(x). \end{aligned}$

Setting $x = a$ ,

$\begin{aligned} 0 &= f''(g(a)) \cdot (g'(a))^2 + f'(g(a)) \cdot g''(a). \end{aligned}$

Making $g''(a)$ the subject,

$\displaystyle g''(a) = -\frac{ f''(g(a)) }{ f'(g(a)) }\cdot (g'(a))^2 .$

Therefore, under the assumption that $f'(g(a)) > 0$ , $g''(a) < 0$ if $f''(g(a)) > 0$ (i.e. $f''$ is convex near $g(a)$ ), and thus $g$ is concave near $a$ .

Question 7. Sketch the graph of $y = 1/x$ for $x \neq 0$ .

(Click for Solution)

Solution. Using Question 3,

$\displaystyle \frac{\mathrm d^2}{\mathrm dx^2}(x^{-1}) = 2x^{-3}.$

For $x > 0$ ,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^{-1}) < 0,\quad \frac{\mathrm d^2}{\mathrm dx^2}(x^{-1}) > 0.$

For $x < 0$ ,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^{-1}) < 0,\quad \frac{\mathrm d^2}{\mathrm dx^2}(x^{-1}) < 0.$

Therefore, we sketch as follows.

Question 8. On the same diagram, sketch the graphs of $y = x^2$ and $y = 1/x^2$ for $x \neq 0$ . What do you notice?

(Click for Solution)

Solution. Using Question 3,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^{-2}) = -2x^{-3},\quad \frac{\mathrm d^2}{\mathrm dx^2}(x^{-2}) = 6x^{-4}.$

For $x > 0$ ,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^{-2}) < 0,\quad \frac{\mathrm d^2}{\mathrm dx^2}(x^{-2}) > 0.$

For $x < 0$ ,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^{-2}) > 0,\quad \frac{\mathrm d^2}{\mathrm dx^2}(x^{-2}) > 0.$

Therefore, we sketch as follows.

Question 9. On the same diagram, sketch the graphs of $y = e^x$ for any $x$ and $y = \ln(x)$ for $x > 0$ . What do you notice?

(Click for Solution)

Solution. Firstly, $e^x > 0$ for any $x$ , so

$\displaystyle \frac{\mathrm d}{\mathrm dx}(e^x) = e^x, \quad \frac{\mathrm d^2}{\mathrm dx^2}(e^x) = e^x$

implies that $y = e^x$ is always convex and increasing.

Secondly, since $e^{\ln(x)} = x$ for $x > 0$ , by Question 6, $y=\ln(x)$ is always concave and increasing. Finally, as inverses, they are reflections about each other about the line $y = x$ , similar to Question 4 and Question 5.

Therefore, we sketch as follows.

—Joel Kindiak, 9 Apr 26, 0111H

KindiakMath

Graphing Techniques

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Graphing Techniques

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