Calculus Question

Question

Given that f(10)=5, f′(10)=1, g(5)=3, and g′(5)=9, what is the

approximate value of g(f(10.01))?

 

Help!?

Answer 1

Using the limit definition of derivative,

 

f'(x) = [f(x + h) - f(x)] / h

g'(x) = [g(x + h) - g(x)] / h         

 

 

Obtain the equations

 

f(10 + h) - f(10)

______________ =  1

         h

 

f(10 + h) - 5 = h

 

 

 
g(5 + h) - g(5)
______________ =  9
         h

 

 

g(5 + h) - 3

____________   = 9

         h

 

g(5 + h) - 3 = 9h

-------------------------------------------------------

 

f(10 + h) = h + 5

 

g(5 + h) = 9h + 3

 

 

These are the equations we will focus on.  Then we have

 

f(10.01) = f(10 + 0.01) = 0.01 + 5 = 5.01

 

 

Now the composition becomes  g(5.01).

 

g(5.01) = g(5 + 0.01) = 9(0.01) + 3 = 3 + 0.09 = 3.09

 

 

g(f(10.01)) = 3.09

Answer 2

 

 f(10.01)  - f(10)

-------------------  = 1    <--- slope = derivative

    10.01 - 10

 

f(10.01) - f(10) = 1( 10.01 - 10)

 

f(10.01) - f(10) = 0.01

 

f(10.01) = f(10) + 0.01

              = 5 + 0.01

              = 5.01

 

--------------------------

 

Likewise,

 

g(5.01) - g(5)

---------------- = 9  

  5.01 - 5

 

g(5.01) - g(5) = 9( 5.01 - 5)

 

g(5.01)- g(5) = 9(0.01)

 

 g(5.01) - 3 = 0.09

 

g(5.01) = 3.09

---------------------------------

 

g(f(10.01)) = g(5.01) = 3.09