## Calculus 8th Edition

a) $\frac{f(x)-f(3)}{x-3}$ b) $f'(3) = \lim\limits_{h \to 0}\frac{f(3+h)-f(3)}{h}$
For part a, we know that the secant line is the average rate of change on the interval between the two points from equation 6. To calculate the slope of this line, we need to calculate the change in y divided by the change in x: $\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$ We can then plug our two given points into the equation to get our answer: $\frac{f(x)-f(3)}{x-3}$ For part b, we know from equation 4 that the equation to find the slope of the tangent line is: $f'(x) = \lim\limits_{h \to 0}\frac{f(a+h)-f(a)}{h}$ The question tells us that point P is located at (3, f(3)). Next we can plug in 3 for a to get the equation for the tangent line: $f'(3) = \lim\limits_{h \to 0}\frac{f(3+h)-f(3)}{h}$