Question

Use this limit to find f′(11) given f(x)=sqrt(x-2)

Answer 1

Answer: f'(11)=1/6

Explanation:

f(x)=sqrt(x-2)

f(x)=(x-2)^1/2

f'(x)=1/2 * (x-2)^(1/2-1)

f'(x)=1/2 * (x-2)^(-1/2)

f'(11)=1/2 * (11-2)^(-1/2)

f'(11)=1/2 * (9)^(-1/2)

f'(11)=1/2 * (9)^(1/2 * (-1))

f'(11)=1/2 * (9^1/2)^(-1)

f'(11)=1/2 * (3)^(-1)

f'(11)=1/2 * 1/3

f'(11)=1*1/(2*3)

f'(11)=1/6

Answer 2

Using the definition of the derivative, we have

`\displaystyle f'(11) = \lim_(x\to11) (√(x-2) - √(11-2))/(x - 11) = \lim_(x\to11) (√(x-2) - 3)/(x - 11)`

Rationalize the numerator.

`\displaystyle (√(x-2)-3)/(x-11) \cdot (√(x-2)+3)/(√(x-2)+3) = (\left(√(x-2)\right)^2 - 3^2)/((x-11)\left(√(x-2)+3\right)) = (x - 11)/((x - 11) \left(√(x-2) + 3\right))`

If
`x\\eq11`, we can simplify the limit to

`\displaystyle f'(11) = \lim_(x\to11) (x-11)/((x-11)\left(√(x-2)+3\right)) = \lim_(x\to11) \frac1{√(x-2) + 3}`

The function in the remaining limit is continuous at
`x=11`, so we end up with

`\displaystyle f'(11) = \frac1{√(11-2) + 3} = \boxed{\frac16}`