Kevin is learning about the basis of calculus and what calculus is actually used for. Unfortunately, Kevin does not understand why calculus is sometimes necessary to find the equation of a line. In Algebra 1, he learned you can find the equation of a line if you are given two points. You find the slope of the line by dividing the up/down difference in the points by the left/right difference, then you use one of the points and the slope to find the y-intercept.

Kevin's teacher, Mr. Banner, offered him extra credit if he could find the slope of a line for the points (4,5) and (4,5) using the method he learned in Algebra 1. Can you see what Mr. Banner did? What is Kevin going to find as he works on those problems?

### Tangents to a Curve

Recall from algebra, if points *P*(*x*_{0},*y*_{0}) and *Q*(*x*_{1},*y*_{1}) are two different points on the curve *y* = *f*(*x*), then the **slope** of the secant line connecting the two points is given by

\begin{align*}m_{sec}\end{align*} | \begin{align*}= \frac {y_1-y_0}{x_1-x_0} = \frac {f(x_1)-f(x_0)}{x_1-x_0}\end{align*} | (1) |
---|

Of course, if we let the point *x*_{1} approach *x*_{o} then *Q* will approach *P* along the graph *f* and thus the slope of the secant line will gradually approach the slope of the tangent line as *x*_{1} approaches *x*_{0}. Therefore, (1) becomes

\begin{align*}m_{sec}\end{align*} | \begin{align*}= \lim_{x_1 \to x_0} \frac {f(x_1)-f(x_0)}{x_1-x_0} \end{align*} | (2) |
---|

To simplify our notation, if we let *h* = *x*_{1} −*x*_{0}, then *x*_{1} = *x*_{0} + *h* and x_{1} → x_{0} becomes equivalent to h → 0. This means that (2) becomes

\begin{align*}m_{sec}\end{align*} | \begin{align*}= \lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h} \end{align*} |
---|

- If the point
*P*(*x*_{0},*y*_{0}) is on the curve*f*, then the tangent line at*P*has a slope that is given by- \begin{align*}m_{tan}=\lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h} \end{align*}
provided that the limit exist. |
---|

Recall that the equation of the tangent line through point (*x*_{0}, *y*_{0}) with slope *m* is the point-slope form of a line: *y* − *y*_{0} = *m*_{tan}(*x* − *x*_{0}).

### Examples

#### Example 1

Earlier, you were given a problem about Kevin, who is having trouble understanding calculus.

Mr. Banner asked Kevin to find the equation of a line given the points (4,5) and (4,5). The points (4, 5) and (4, 5) are the same, so the \begin{align*}\frac{rise}{run}\end{align*} would be \begin{align*}\frac{0}{0}\end{align*} - Kevin was just introduced to the need for differential calculus!

#### Example 2

Find line tangent to the curve *f*(*x*) = *x*^{3} that passes through point *P* (2,8).

Since *P*(*x*_{0}, *y*_{0}) = (2, 8), using the slope of the tangent equation we have

\begin{align*}m_{tan}\end{align*} | \begin{align*}= \lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h} \end{align*} | ||
---|---|---|---|

and we get | |||

\begin{align*}m_{tan}\end{align*} | \begin{align*}= \lim_{h \to 0} \frac {f(2+h)-f(2)}{h} \end{align*} | ||

\begin{align*}= \lim_{h \to 0} \frac {(h^3+6h^2+12h+8)-8}{h} \end{align*} | |||

\begin{align*}= \lim_{h \to 0} \frac {h^3+6h^2+12h}{h} \end{align*} | |||

\begin{align*}= \lim_{h \to 0} (h^2+6h+12) \end{align*} | |||

\begin{align*}= 12\end{align*} |

Thus the slope of the tangent line is 12. Using the point-slope formula above, we find that the equation of the tangent line is *y* - 8 = 12 (*x* - 2) or *y* = 12*x* - 16.

#### Example 3

If *f*(*x*) = *x*^{2} − 3,find *f'* (*x*) and use the result to find the slope of the tangent line at *x* = 2 and *x* = −1.

Since \begin{align*}f'(x)= \lim_{h \to 0} \frac {f(x+h)-f(x)}{h} \end{align*} then

\begin{align*}f'(x)\end{align*} | \begin{align*}= \lim_{h \to 0} \frac {\left [ {(x+h)^2-3} \right ]- \left [ {x^2-3} \right ]}{h} \end{align*} | ||
---|---|---|---|

\begin{align*}= \lim_{h \to 0}\frac {x^2+2xh+h^2-3-x^2+3}{h}\end{align*} | |||

\begin{align*}= \lim_{h \to 0} \frac {2xh+h^2}{h}\end{align*} | |||

\begin{align*}= \lim_{h \to 0} (2x+h)\end{align*} | |||

\begin{align*}= 2x\end{align*} |

To find the slope, we simply substitute *x* = 2 into the result *f'* (*x*):

f'(x) |
= 2x |
||
---|---|---|---|

f'(2) | =2(2) | ||

= 4 | |||

and | |||

f'(x) | =2x | ||

f'(-1) | =2(-1) | ||

= -2 |

Thus slope of the tangent line at *x* = 2 and *x* = −1 are 4 and −2 respectively.

#### Example 4

Find the slope of the tangent line to the curve *y* = 1/*x* that passes through the point (1, 1).

Using the slope of the tangent formula,

\begin{align*}f'(x)\end{align*} | \begin{align*}= \lim_{h \to 0} \frac {f(x+h)-f(x)}{h}\end{align*} | ||
---|---|---|---|

and substituting \begin{align*}y=\frac{1}{x}\end{align*} | |||

\begin{align*}f'(x)\end{align*} | \begin{align*}= \lim_{h \to 0} \frac {\left ( \frac{1}{x+h} \right )-\frac {1}{x}}{h}\end{align*} | ||

\begin{align*}= \lim_{h \to 0} \frac {\frac {x-x-h}{x \left ({x+h} \right)}}{h}\end{align*} | |||

\begin{align*}= \lim_{h \to 0} \frac {x-x-h}{hx \left ({x+h} \right)}\end{align*} | |||

\begin{align*}= \lim_{h \to 0} \frac {-h}{hx \left ({x+h} \right)}\end{align*} | |||

\begin{align*}= \lim_{h \to 0} \frac {-1}{x \left ({x+h} \right)}\end{align*} | |||

\begin{align*}= \frac {-1}{x^2}\end{align*} | |||

For x = 1, the slope is |
|||

\begin{align*}f'(x)\end{align*} | \begin{align*}= \frac {-1}{1} = -1\end{align*} | ||

\begin{align*}= -1\end{align*} |

Thus the slope of the tangent line at *x* = 1 for the curve *y* = 1/*x* is *m* = −1. To find the equation of the tangent line, we simply use the point-slope formula,

y - y_{0} |
= m(x - x_{0}) |
||
---|---|---|---|

Where (x_{0}, y_{0}) = (1, 1). |
|||

y - 1 |
= -1(x - 1) |
||

y |
= -x + 1 + 1 |
||

y |
= -x + 2 |

So the equation of the tangent line is *y* = -*x* + 2.

#### Example 5

Given the function \begin{align*}y = \frac{1}{2} x^2\end{align*} and the values of \begin{align*}x_0 = 3\end{align*} and \begin{align*}x_1 = 4\end{align*}, find:

- The average rate of change of
*y*with respect to*x*over the interval [*x*_{0},*x*_{1}].

Identify the two points by substituting 3 and 4 in for *x* in the function \begin{align*}f(x) = \frac{1}{2}x^2\end{align*}

Substitute the two points (3, 4.5) and (4, 8) into the average rate of change formula: \begin{align*}m = \frac{y_1 - y_0}{x_1 - x_0}\end{align*}

Average rate of change = \begin{align*}\frac{7} {2}\end{align*}

- The slope of the secant line connecting
*x*_{0}and*x*_{1.}

The slope of the secant line between \begin{align*}x_0\end{align*} and \begin{align*}x_1\end{align*} is the slope between \begin{align*}(3, 4.5)\end{align*} and \begin{align*}(4, 8)\end{align*}, which is \begin{align*}\frac{7} {2}\end{align*}.

- The instantaneous rate of change of
*y*with respect to*x*at*x*_{0}.

Instantaneous rate of change is the slope *at* x = 3.

Use the formula: \begin{align*}\frac{f(x + h) - f(x)}{h}\end{align*} where \begin{align*}f(x) = \frac{1}{2}x^2\end{align*} and \begin{align*}x = 3\end{align*}

\begin{align*}\frac{f(3 + h) - f(3)}{h}\end{align*} ..... Substitute 3 for x

\begin{align*}\frac{\frac{1}{2}(3 + h)^2 - \frac{1}{2}(3)^2}{h}\end{align*} ..... Replace \begin{align*}f(x) \to \frac{1}{2}x^2\end{align*}

\begin{align*}\frac{\frac{1}{2}(9) + \frac{1}{2}(6h) + \frac{1}{2}h^2 - \frac{1}{2}9}{h}\end{align*} ..... FOIL and Distribute the 1/2

\begin{align*}\frac{6h + h^2}{2h}\end{align*} ..... Simplify

\begin{align*}3+\frac{h}{2}\end{align*} ..... Simplify again

\begin{align*}3\end{align*} ..... As \begin{align*}h \to 0\end{align*}

\begin{align*}\therefore\end{align*} the instantaneous slope at x = 3 is 3

- The slope of the tangent line at
*x*_{1}.

The slope of the tangent at 4 is the same as the instantaneous rate of change at \begin{align*}x = 4\end{align*}

This is the same series of steps as with x = 3 above

\begin{align*}\therefore\end{align*} the slope at x = 4 is 4

#### Example 6

Given the function \begin{align*}f(x) = \frac{1}{x}\end{align*} and the values \begin{align*}x_{0} = 2\end{align*} and \begin{align*}x_1 = 3\end{align*}, find:

- The average rate of change of
*y*with respect to*x*over the interval [*x*_{0},*x*_{1}].

Identify the two points by substituting 2 and 3 in for *x* in the function \begin{align*}f(x) = \frac{1}{x}\end{align*} to get \begin{align*}(2, \frac{1}{2}) | (3, \frac{1}{3})\end{align*}

Substitute the two points \begin{align*}(2, \frac{1}{2}) | (3, \frac{1}{3})\end{align*} into the average rate of change formula: \begin{align*}m = \frac{y_1 - y_0}{x_1 - x_0}\end{align*}

Average rate of change = \begin{align*}\frac{-1}{6}\end{align*}

- The slope of the secant line connecting
*x*_{0}and*x*_{1}.

The slope of the secant line between \begin{align*}x_0\end{align*} and \begin{align*}x_1\end{align*} is the slope between \begin{align*}(2, \frac{1}{2})\end{align*} and \begin{align*}(3, \frac{1}{3})\end{align*}, which is \begin{align*}\frac{-1}{6}\end{align*}.

- The instantaneous rate of change of
*y*with respect to*x*at*x*_{0}.

Instantaneous rate of change at *x*_{0} is the slope *at* *x*= 2.

Use the formula: \begin{align*}\frac{f(x + h) - f(x)}{h}\end{align*} where \begin{align*}f(x) = \frac{1}{x}\end{align*} and \begin{align*}x = 2\end{align*}

\begin{align*}\frac{f(2 + h) - f(2)}{h}\end{align*} ..... Substitute 2 for x

\begin{align*}\frac{\frac{1}{2 + h} - \frac{1}{2}}{h}\end{align*} ..... Replace \begin{align*}f(x) \to \frac{1}{x}\end{align*}

\begin{align*}(\frac{1}{2 + h} - \frac{1}{2}) \cdot \frac{1}{h}\end{align*} ..... We had a fraction divided by a fraction, invert to multiply

\begin{align*}\frac{(2)(1)}{2(2 + h)} - \frac{(2 + h)(1)}{2(2 + h)} \cdot \frac{1}{h}\end{align*} ..... Set common denominators

\begin{align*}\frac{(2) - (2 + h)}{(2 + h)(2)(h)}\end{align*} ..... Simplify

\begin{align*}\frac{-h}{4h +2h^2}\end{align*} ..... Simplify again

\begin{align*}\frac{-1}{4 + 2h}\end{align*} ..... once more (canceling the *h*)

\begin{align*}\frac{-1}{4}\end{align*} ..... As \begin{align*}h \to 0\end{align*}

\begin{align*}\therefore\end{align*} the instantaneous slope at x = 2 is \begin{align*}\frac{-1} {4}\end{align*}

- The slope of the tangent line at
*x*_{1}.

The slope of the tangent at 3 is the same as the instantaneous rate of change at \begin{align*}x = 3\end{align*}

This is the same series of steps as with x = 2 above

\begin{align*}\therefore\end{align*} the slope at x = 3 is \begin{align*}\frac{-1} {9}\end{align*}

### Review

- What is the line connecting two points \begin{align*}(x_0, y_0)\end{align*} and \begin{align*}(x_1, y_1)\end{align*} on a curve called?
- As \begin{align*}(x_0, y_0)\end{align*} gets immeasurably close to \begin{align*}(x_1, y_1)\end{align*} the term describing the line between them becomes: "the ____________ line"
- The expression \begin{align*}f(x_0 + h) - f(x_0)\end{align*} is used to describe what distance in the process of finding the slope of a tangent line?
- When calculating the slope of a tangent, what value is assumed to go to 0 as the two chosen points get closer and closer?
- What does the concept of limit, discussed in prior lessons, have to do with finding the slope of a line tangent to a curve?

Find the equation of the tangent line:

- What is the equation of the tangent line at \begin{align*}x = -3\end{align*} assuming that \begin{align*}r(-3) = - 5\end{align*} and \begin{align*}r'(-3) = 1\end{align*}?
- What is the equation of the tangent line at \begin{align*} x = 1\end{align*} assuming that \begin{align*}r(1) = 3\end{align*} and \begin{align*}r'(1) = -5\end{align*}?
- What is the equation of the tangent line at \begin{align*}x = 2\end{align*} assuming that \begin{align*}g(2) = 1\end{align*} and \begin{align*}g'(2) = -3\end{align*}?
- What is the equation of the tangent line at \begin{align*}x = 4\end{align*} assuming that \begin{align*}u(4) = 4\end{align*} and \begin{align*}u'(4) = 3\end{align*}?
- What is the equation of the tangent line at \begin{align*}x = -4\end{align*} assuming that \begin{align*}t(-4) = 2\end{align*} and \begin{align*}t'(-4) = 5\end{align*}?

Find the equation of the tangent line:

- Find the equation of the tangent line to the graph of \begin{align*}h(x) = -5x^3 - 3x^2 + x + 3 \end{align*} at \begin{align*}x = 1\end{align*}
- Find the equation of the tangent line to the graph of \begin{align*}t(x) = -2x\end{align*} at \begin{align*}x = -2\end{align*}
- Find the equation of the tangent line to the graph of \begin{align*}m(x) = 3x^3 + 3x^2 + 4x + 4\end{align*} at \begin{align*}x = 1\end{align*}
- Find the equation of the tangent line to the graph of \begin{align*}q(x) = -x^3 - 4x^2 + 4x + 3\end{align*} at \begin{align*}x = -2\end{align*}
- Find the equation of the tangent line to the graph of \begin{align*}t(x) = -4x^2 + 2x - 4\end{align*} at \begin{align*}x = -1\end{align*}
- Find the equation of the tangent line to the graph of \begin{align*}h(x) = -4x^3 + 2x^2 - 3x + 3\end{align*} at \begin{align*}x = -1\end{align*}
- Find the equation of the tangent line to the graph of \begin{align*}m(x) = x \end{align*} at \begin{align*}x = 0\end{align*}
- Find the equation of the tangent line to the graph of \begin{align*}s(x) = -3x^2 - 2x + 3 \end{align*} at \begin{align*}x = 0\end{align*}
- Find the equation of the tangent line to the graph of \begin{align*}c(x) = -3\end{align*} at \begin{align*}x = 0\end{align*}
- Find the equation of the tangent line to the graph of \begin{align*}b(x) = -5x^4 + 3x^3 - x^2 + 5x - 3 \end{align*} at \begin{align*}x = -1\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 8.7.