Question

Using mean value theorem find the √11

Answer 1

Let $y=f(x)=\sqrt{x}.$

The mean value theorem states that

$$f(b) - f(a) = f'(c)(b-a), \quad a<c<b$$

Let $b$ is near $a$ then we can write $\Delta x=b-a$ and rewrite the formula as

$$\Delta y= f'(c)\Delta x$$

If $\Delta x \to 0,$ then we obtain

$$dy=f'(x)dx$$

Thus, $f(x+\Delta x) - f(x) \approx f'(x)dx.$

Therefore, $f(x+\Delta x) \approx f(x) + f'(x)dx.$

In this case,

$$x+\Delta x=11, x = 9, \Delta x=dx=2, \text{ and } f'(x) = \frac{1}{2\sqrt{x}}$$

$f(11)= \sqrt{11} \approx \sqrt{9} + \frac{1}{2\sqrt{9}}\cdot 2 = 3 + \frac{1}{3} \approx 3.3333$

Answer: $\sqrt{11} \approx 3.3333.$