Textbook Problem

In Example 6 in this section, we were given

$f\left(x\right)=3x^2+2x+11\text{ and found }f\text{'}\left(x\right)=6x+2$. Find

(a) the instantaneous rate of change of f(x) at x = 6.

(b) the slope of the tangent to the graph of y = f(x) at x = 6.

(c) the point on the graph of y = f(x) at x = 6.

EXAMPLE 6 Tangent Line

Given $y=f(x)=3x^2+2x +11$, find

(a) the derivative of f(x) at any point (x, f(x)).

(b) the slope of the tangent to the curve at (1, 16).

(c) the equation of the line tangent to $y=3x^2+2x+11$ at (1.16).

Solution

(a) The derivative of f(x) at any value x is denoted by f'(x) and is

$\begin{array}{l}y\text{'}=f\text{'}(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left[3{\left(x+h\right)}^2+2\left(x+h\right)+11\right]-\left(3x^2+2x+11\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{3\left(x^2+2xh+h^2\right)2x+2h+11-3x^2-2x-11}{h}\end{array}$

$=\underset{h\to 0}{\lim }\frac{6xh+3h^2+2h}{h}=\underset{h\to 0}{\lim }\left(6x+3h+2\right)=6x+2$

(b) The derivative is $f\text{'}\left(x\right)=6x+2$, so the slope of the tangent to the curve at (1,16) is $f\text{'}\left(1\right)\text{ = }6\left(1\right)+2=8$.

(c) The equation of the tangent line uses the given point (1, 16) and the slope m = 8. Using $y-y_1=m(x-x_1)$ gives $y-16=8\left(x-\text{1}\right),\text{ or }y=8x+8$.

The discussion in this section indicates that the derivative of a function has several interpretations.