#### Answer

$$\frac{8}{3}$$

#### Work Step by Step

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.