Math Problem Statement

Using the limit definition of the derivative, find the value of the derivative of f(x) = 2/(x+5) at the point where x = 4. Find the slope-intercept form of the equation of the tangent line at x = 4.

Solution

Let's solve the problem step-by-step:

Step 1: Use the limit definition of the derivative

The limit definition of the derivative is given by:

f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Here, f(x)=2x+5f(x) = \frac{2}{x+5}. So let's apply the limit definition to this function.

Step 2: Set up the difference quotient

We need to calculate the difference quotient for f(x)f(x):

f(x)=limh02(x+h)+52x+5hf'(x) = \lim_{h \to 0} \frac{\frac{2}{(x+h)+5} - \frac{2}{x+5}}{h}

Simplify the numerator:

=2x+h+52x+5=2(x+5)2(x+h+5)(x+h+5)(x+5)= \frac{2}{x+h+5} - \frac{2}{x+5} = \frac{2(x+5) - 2(x+h+5)}{(x+h+5)(x+5)}

=2(x+5)2(x+h+5)(x+h+5)(x+5)=2x+102x2h10(x+h+5)(x+5)= \frac{2(x+5) - 2(x+h+5)}{(x+h+5)(x+5)} = \frac{2x + 10 - 2x - 2h - 10}{(x+h+5)(x+5)}

=2h(x+h+5)(x+5)= \frac{-2h}{(x+h+5)(x+5)}

Step 3: Divide by hh and take the limit

Now divide by hh:

f(x)=limh02hh(x+h+5)(x+5)=limh02(x+h+5)(x+5)f'(x) = \lim_{h \to 0} \frac{-2h}{h(x+h+5)(x+5)} = \lim_{h \to 0} \frac{-2}{(x+h+5)(x+5)}

As h0h \to 0, this becomes:

f(x)=2(x+5)2f'(x) = \frac{-2}{(x+5)^2}

Step 4: Calculate the derivative at x=4x = 4

Now substitute x=4x = 4 into the derivative:

f(4)=2(4+5)2=292=281f'(4) = \frac{-2}{(4+5)^2} = \frac{-2}{9^2} = \frac{-2}{81}

So, the slope of the tangent line at x=4x = 4 is 281\frac{-2}{81}.

Step 5: Find the equation of the tangent line

The slope-intercept form of the tangent line is:

yf(4)=f(4)(x4)y - f(4) = f'(4)(x - 4)

First, find f(4)f(4):

f(4)=24+5=29f(4) = \frac{2}{4+5} = \frac{2}{9}

Now, use the slope 281\frac{-2}{81} and the point (4,29)(4, \frac{2}{9}):

y29=281(x4)y - \frac{2}{9} = \frac{-2}{81}(x - 4)

Simplify this to get the equation in slope-intercept form:

y=281x+881+29y = \frac{-2}{81}x + \frac{8}{81} + \frac{2}{9}

y=281x+881+1881y = \frac{-2}{81}x + \frac{8}{81} + \frac{18}{81}

y=281x+2681y = \frac{-2}{81}x + \frac{26}{81}

So the equation of the tangent line is:

y=281x+2681y = \frac{-2}{81}x + \frac{26}{81}

Would you like further details or have any questions?

5 Related Questions:

  1. How is the limit definition of the derivative applied to different types of functions?
  2. What is the geometric meaning of the derivative in this context?
  3. How would the derivative of f(x)=2x+5f(x) = \frac{2}{x+5} change if the function was different, say f(x)=3x+2f(x) = \frac{3}{x+2}?
  4. How do you compute the derivative of more complex rational functions?
  5. What other methods can be used to find the derivative apart from the limit definition?

Tip:

For complex fractions, always simplify the numerator carefully before taking the limit to avoid errors.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Limit Definition of Derivative
Slope of Tangent Line
Slope-Intercept Form

Formulas

f'(x) = lim_{h -> 0} [(f(x+h) - f(x))/h]
y - f(a) = f'(a)(x - a)

Theorems

Limit Definition of Derivative

Suitable Grade Level

College