Answer:
15
Step-by-step explanation:
Let f(x) be the function
[tex]\bf f(x)=\sqrt{x}[/tex]
A linear approximation of f is the Taylor polynomial of degree one:
[tex]\bf f(x)\approx f(a)+f'(a)(x-a)[/tex]
Taking a = 16, and given that
[tex]\bf f'(x)=\displaystyle\frac{1}{2\sqrt{x}}[/tex]
we get
[tex]\bf f(x)\approx f(16)+f'(16)(x-16)=\sqrt{16}+\displaystyle\frac{1}{2\sqrt{16}}(x-16)=4+\displaystyle\frac{x-16}{8}[/tex]
so
[tex]\bf \sqrt{14}=f(14)\approx 4+\displaystyle\frac{14-16}{8}=3.75\Rightarrow\\\\\Rightarrow 4\sqrt{14}\approx 4(3.75)=15[/tex]
Since 16 > 14, we can deduce that this is an overestimate.
