Calculus 1 Question

Question

Use differentials to approximate the value of the expression. Compare your answer with that of a calculator. (Round your answers to four decimal places.)

(5.99)^3

 

Answer 1

You need to find the linear approximation.

 

Suppose you had a function such that

 

f(x) = x³

 

This function was derived using the expression given in the problem.  Since you are taking some input and cubing it.

 

Lets write down the linear approximation formula so we can see what we need to find.

 

L(x) = f(x_o) + f'(x₀)(x - x₀)

 

 

We need to find the derivative of f(x).

 

f'(x) = 3x²

 

Since x=5.99 is close to x=6, we use x₀=6.

 

f'(6) = 3(6)²

       = 3 * 36

       = 108

 

f(6) = 6³

      = 216

 

 

Now we plug in these value of f(x₀) and f'(x_o) into the approximation.

 

L(x) = 216 + 108(x - 6)

L(x) = 216 + 108x - 648

L(x) = 108x - 432