Chapter 12, Problem 6CT

a.

To determine

find the slope at the given point.

Answer to Problem 6CT

The slope of the tangent to $f\left(x\right)=3x^2-5x-2$ at $\left(2,0\right)$ is $\boxed{7}$

Explanation of Solution

Given information:

Find a formula for the slope of the graph of $f$ at the point $\left(x,f\left(x\right)\right)$ . Then use the formula to find the slope at the given point.

  $f\left(x\right)=3x^2-5x-2$

  $\left(2,0\right)$

Calculation:

Consider the given function,

  $f\left(x\right)=3x^2-5x-2$

As both the points are lying on the curve, the line joining these two points would be the secant of the curve.

If $\Delta x$ approaches zero, then secant tends to become the tangent to the curve at point $X$

So that slope of tangent at $X$ is

  $m_{\tan }=\underset{h\to \infty }{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

The expression is defined as the derivative of the function.

Now find slope of the tangent to the curve at the defined point, by using standard formula of slope.

  $\begin{array}{l}{m}_{\tan }=\underset{h\to \infty }{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ f\left(x\right)=3x^2-5x-2\\ f\left(x+h\right)=3{\left(x+h\right)}^2-5\left(x+h\right)-2\end{array}$

Apply algebraic identity,

  ${\left(a+b\right)}^2=a^2+b^2+2ab$

  $\begin{array}{l}f\left(x+h\right)=3x^2+h^2+6xh-5x-5h-2\\ =3x^2-5x-2+h^2+6xh-5h\end{array}$

Now find the difference between $f\left(x+h\right)$ and $f\left(x\right)$ .

  $\begin{array}{l}f\left(x+h\right)-f\left(x\right)\\ =3x^2-5x-2+h^2+6xh-5h-\left(3x^2-5x-2\right)\\ =6xh+h^2-5\\ =h\left(6x+h-5\right)\end{array}$

Now divide the difference by $h$ .

  $\frac{f\left(x+h\right)-f\left(x\right)}{h}$

  $\begin{array}{l}=\frac{\cancel{h}\left(6x+h-5\right)}{\cancel{h}}\\ =6x+h-5\end{array}$

Now apply limit $\underset{h\to 0}{\lim }$ we get,

  $\begin{array}{l}{m}_{\tan }=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }6x+h-5\\ =6x-5\end{array}$

Hence the slope of the tangent for any point for defined function is

  $\boxed{6x-5}$

Now substitute $x=2$

  $\begin{array}{l}{m}_{\tan }=6x-5\\ =6\left(2\right)-5\\ =12-5\\ =7\end{array}$

Hence the slope of the tangent to $f\left(x\right)=3x^2-5x-2$ at $\left(2,0\right)$ is $\boxed{7}$

b.

To determine

find the slope at the given point.

Answer to Problem 6CT

The slope of the tangent to $f\left(x\right)=2x^3+6x$ at $\left(-1,-8\right)$ is $\boxed{12}$

Explanation of Solution

Given information:

Find a formula for the slope of the graph of $f$ at the point $\left(x,f\left(x\right)\right)$ . Then use the formula to find the slope at the given point.

  $f\left(x\right)=2x^3+6x$

  $\left(-1,-8\right)$

Calculation:

Consider the given function,

  $f\left(x\right)=2x^3+6x$

As both the points are lying on the curve, the line joining these two points would be the secant of the curve.

If $\Delta x$ approaches zero, then secant tends to become the tangent to the curve at point $X$

So that slope of tangent at $X$ is

  $m_{\tan }=\underset{h\to \infty }{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

The expression is defined as the derivative of the function.

Now find slope of the tangent to the curve at the defined point, by using standard formula of slope.

  $\begin{array}{l}f\left(x\right)=2x^3+6x\\ f\left(x+h\right)=2{\left(x+h\right)}^3+6\left(x+h\right)\end{array}$

Apply algebraic identity,

  ${\left(a+b\right)}^3=a^3+b^3+3a{b}^2+3a^{2}b$

  $\begin{array}{l}f\left(x+h\right)=2\left(x^3+h^3+3x^{2}h+3x{h}^2\right)+6x+6h\\ =2x^3+6x^{2}h+6x{h}^2+2h^3+6x+6h\end{array}$

Now find the difference between $f\left(x+h\right)$ and $f\left(x\right)$ .

  $\begin{array}{l}f\left(x+h\right)-f\left(x\right)\\ =2x^3+6x^{2}h+6x{h}^2+2h^3+6x+6h-\left(2x^3+6x\right)\\ =6x^{2}h+6x{h}^2+2h^3+6h\\ =h\left(6x^2+6xh+2h^2+6\right)\end{array}$

Now divide the difference by $h$ .

  $\frac{f\left(x+h\right)-f\left(x\right)}{h}$

  $\begin{array}{l}\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\frac{h\left(6x^2+6xh+2h^2+6\right)}{h}\\ =\left(6x^2+6xh+2h^2+6\right)\end{array}$

Now apply limit $\underset{h\to 0}{\lim }$ we get,

  $\begin{array}{l}{m}_{\tan }=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }6x^2+6xh+2h^2+6\\ =6x^2+0+0+6\\ m_{\tan }=6x^2+6\end{array}$

Hence the slope of the tangent for any point for defined function is

  $m_{\tan }=6x^2+6$

Now substitute $x=-1$

  $\begin{array}{l}{m}_{\tan }=6x^2+6\\ =6{\left(-1\right)}^2+6\\ =6+6\\ =12\end{array}$

Hence the slope of the tangent to $f\left(x\right)=2x^3+6x$ at $\left(-1,-8\right)$ is $\boxed{12}$