Chapter 2.3, Problem 84E

Solution

a.

To show that the slope of the line through the points $P(a,f(a))$ and $Q(-a,f(-a))$ is $a^{n-1}$ where $f\left(x\right)=x^n$ .

Given the function:

  $f\left(x\right)=x^n$

Formula used:

The slope $m$ of the line passing through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ :

  $m=\frac{y_2-y_1}{x_2-x_1}$ .

Proof:

The slope $m$ of the line passing through the points $P(a,f(a))$ and $Q(-a,f(-a))$ :

  $\begin{array}{l}m=\frac{f\left(-a\right)-f\left(a\right)}{\left(-a\right)-a}\\ m=\frac{{\left(-a\right)}^n-a^n}{-2a}\\ m=\frac{{\left(-1\right)}^n{a}^n-a^n}{-2a}\\ m=\frac{{\left(-1\right)}^n{a}^{n-1}-a^{n-1}}{-2}\\ m=\frac{{\left(-1\right)}^{n-1}{a}^{n-1}+a^{n-1}}{2}\end{array}$ .

Now, since $n$ is odd, $n-1$ is even.

Then, it follows that ${\left(-1\right)}^{n-1}=1$ .

Now, the slope:

  $\begin{array}{l}m=\frac{{\left(-1\right)}^{n-1}{a}^{n-1}+a^{n-1}}{2}\\ m=\frac{2a^{n-1}}{2}\\ m=a^{n-1}\end{array}$ .

b.

To determine the equation of the line passing through the point $\left(x_0,f\left(x_0\right)\right)$ that has the slope $a^{n-1}$ where $x_0=a^{\frac{1}{n-1}}$ .

The equation of the line passing through the point $\left(x_0,f\left(x_0\right)\right)$ that has the slope $a^{n-1}$ where $x_0=a^{\frac{1}{n-1}}$ :

  $y=a^{n-1}x+\left(a^{\frac{n}{n-1}}-1\right)$ .

Concept Used:

The equation of the line passing through the point $\left(x_0,y_0\right)$ having the slope $m$ :

  $y-y_0=m\left(x-x_0\right)$ .

Find the equation of the line passing through the point $\left(x_0,f\left(x_0\right)\right)$that has the slope$a^{n-1}$where $x_0=a^{\frac{1}{n-1}}$:

The equation of the line passing through the point $\left(x_0,f\left(x_0\right)\right)$ that has the slope $a^{n-1}$ :

  $\begin{array}{l}y-f\left(x_0\right)=a^{n-1}\left(x-x_0\right)\\ y=a^{n-1}x+\left(f\left(x_0\right)-a^{n-1}{x}_0\right)\end{array}$ .

Substitute $x_0=a^{\frac{1}{n-1}}$ in the equation $y=a^{n-1}x+\left(f\left(x_0\right)-a^{n-1}{x}_0\right)$ :

  $\begin{array}{l}y=a^{n-1}x+\left(f\left(a^{\frac{1}{n-1}}\right)-a^{n-1}{a}^{\frac{1}{n-1}}\right)\\ y=a^{n-1}x+\left({\left(a^{\frac{1}{n-1}}\right)}^n-1\right)\\ y=a^{n-1}x+\left(a^{\frac{n}{n-1}}-1\right)\end{array}$ .

Thus, the equation of the line is $y=a^{n-1}x+\left(a^{\frac{n}{n-1}}-1\right)$ .

Conclusion:

The equation of the line passing through the point $\left(x_0,f\left(x_0\right)\right)$ that has the slope $a^{n-1}$ where $x_0=a^{\frac{1}{n-1}}$ :

  $y=a^{n-1}x+\left(a^{\frac{n}{n-1}}-1\right)$ .

c.

To show the graph of the line $y=a^{n-1}x+\left(a^{\frac{n}{n-1}}-1\right)$ and that of $f\left(x\right)=x^n$ when $n=3$ and $a=3$ in the same window with dimension $[-5,5]$ and $[-30,30]$ .

To find the equation of the line when $n=3$ and $a=3$ , substitute $n=3$ and $a=3$ in the equation $y=a^{n-1}x+\left(a^{\frac{n}{n-1}}-1\right)$ :

  $\begin{array}{l}y=3^{3-1}x+\left(3^{\frac{3}{3-1}}-1\right)\\ y=3^{2}x+\left(3^{\frac{3}{2}}-1\right)\\ y=9x+\left(3^{\frac{3}{2}}-1\right)\end{array}$ .

Thus, the equation of the line when $n=3$ and $a=3$ is $y=9x+\left(3^{\frac{3}{2}}-1\right)$ .

To find the function when $n=3$ and $a=3$ , substitute $n=3$ in $f\left(x\right)=x^n$ :

  $f\left(x\right)=x^3$ .

Thus, the function $f$ when $n=3$ is $f\left(x\right)=x^3$ .

Graph the function $f\left(x\right)=x^3$ and the line $y=9x+\left(3^{\frac{3}{2}}-1\right)$ in a window with dimension $[-5,5]$ by $[-30,30]$:

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