Difference Quotient and Equation of the Tangent Line

Question

Calculate the difference quotient for f(x)=1/(√2x+5) at a=4.

f(4+h)-f(4)/h = ?

f'(4) = -1/13^(3/2)

Find an equation of the tangent line to f(x) at a=4

What would the difference quotient and equation of the tangent line be?

Answer 1

The difference quotient is simply the slope between two points, the first is (x, f(x)) or, in this case (4, f(4)), and the second is (x + h, f(x + h)). However, in this case, based on the other information you provided, I believe you are looking for the limit of the difference quotient as h approaches zero. That is known as the derivative:

To determine what this turns out to be, just plug in the associated elements:

f(x + h) = f(4 + h) = 1/√(2(4+h)+5) = 1/√(2h + 13)

and f(x) = f(4) = 1/√(2(4)+5) = 1/√13

So [f(4 + h) - f(4)]/h = [1/√(2h + 13) - 1/√13]/h

There is a lot of simplifying to get it to turn into -1/13^(3/2) however it does. To do so, working with only f(4 + h) - f(4):

1) Get a common denominator and combine/simplify

2) Rationalize the denominator

Put together the whole difference quotient by dividing by "h"

3) Multiply the numerator denominator by the conjugate of the numerator

4) Cancel out an "h" and take the limit by plugging in h = zero.

5) Simplify

So now you have f '(4) = -1/13^(3/2). That is the slope at the point (4, f(4)) or (4, 1/√13) or (4, √13/13).

Since you have the slope and a point, you can use the point-slope form of a line: y - y₁ = m(x - x₁) where m = -1/13^(3/2) and (x₁, y₁) = (4, √13/13)