## Calculus (3rd Edition)

$$\frac{8}{3}$$
For $h=0.3$, and from the given figure, we have $f(1)= 1.2$ and $f(0.7)=2$. Thus: \begin{align*} \frac{f(0.7+h)-f(0.7)}{h}&=\frac{f(1)-f(0.7)}{0.3} \\ &\approx \frac{2.8-2}{0.3}\\ &=\frac{8}{3} \end{align*} This value (the difference quotient) is larger than $f'(0.7)'$. We can tell from the graph that the slope of the tangent line will be smaller than the slope of the secant line at the given point.