Chapter 4 - Differentiation - 4.1 Linear Approximation and Applications - Exercises - Page 173: 25

$$f(4.03 )=2.01$$

From the given figure, we see that $f(4)=2$ and \begin{align*} f^{\prime}(4)&=\frac{4-2}{10-4}\\ &=\frac{1}{3} \end{align*} Then we have \begin{align*} \Delta f&=f(a+\Delta x)-f(a)\\ f(4.03)&=\Delta f+ f(4) \\ &=2+\frac{1}{3}(0.03)\\ &=2.01 \end{align*}