Science:Math Exam Resources/Courses/MATH110/April 2018/Question 08 (b)/Solution 1

We use the formula ${\textstyle f(x)\approx f(a)+f'(a)(x-a)}$, as given in the hint.

First observe that by the power rule we have ${\textstyle f'(x)=0.5x^{-0.5}}$. Thus, using ${\textstyle a=1}$, we have ${\textstyle f(a)={\sqrt {1}}=1}$ and ${\textstyle f'(a)=0.5\cdot 1^{-0.5}=0.5}$. Plugging these into the formula gives

$$f(x)\approx 1+0.5(e-1)=0.5e+0.5={\frac {e+1}{2}}.$$

Answer: The linear approximation gives ${\textstyle {\color {blue}e^{0.5}\approx {\frac {e+1}{2}}\approx 1.859}}$.