Chapter 11, Problem 78RE

To determine

To find: the slope of the graph, equation of tangent line and graph the function.

Answer to Problem 78RE

Theslopes of the graph is 11 and equation of the tangent line is $y=11x-16$ .

Explanation of Solution

Given:

  $f\left(x\right)=x^3-x$

Calculation:

Slope of the graph,

  $\begin{array}{l}f\left(x\right)=x^3-x\\ m=\underset{h\to 0}{\lim }\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{\left(x+h\right)\left({\left(x+h\right)}^2-1\right)-\left(x^3-x\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{\left(x+h\right)\left(x^2+h^2+2xh-1\right)-x^3+x}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{x^3+h^{2}x+2x^{2}h-x+x^{2}h+h^3+2x{h}^2-h-x^3+x}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{h^{2}x+2x^{2}h+x^{2}h+h^3+2x{h}^2-h}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{h\left(hx+2x^2+x^2+h^2+2xh-1\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(h^2+3hx+3x^2-1\right)\\ =0^2+3\left(0\right)x+3x^2-1\\ =3x^2-1\end{array}$

At point $\left(2,6\right)$ ,

  $m=3{\left(2\right)}^2-1=11$

Using point slope formula,

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-6=11\left(x-2\right)\\ y-6=11x-22\\ y=11x-16\end{array}$

Conclusion:

Therefore,the slopes of the graph is 11 and equation of the tangent line is $y=11x-16$ .