Answer
$$ 0.05,\ \ 0.000610,\ \ 1.24\%$$
Work Step by Step
Given $$f(x)=\sqrt{1+x}, \quad a=3, \quad \Delta x=0.2$$ Since $$ f'(x) = \frac{1}{2\sqrt{1+x}} ,\ \ \ f'(3)= 0.25 $$ Then \begin{align*} \Delta f \approx f^{\prime}(a) \Delta x&=(0.25)(0.2)\\ &=0.05 \end{align*} Since the change $\Delta f$ given by \begin{align*} \Delta f&=f(a+\Delta x)-f(a)\\ &=f(3.2)-f(3)\\ &=\sqrt{4.2}-2\\ & \approx 0.049390 \end{align*} to find error $$|0.049390-0.05|=0.000610$$ and the percentage $$\frac{0.000610}{0.049390} \times 100 \% \approx 1.24 \%$$
