Let f (x) = 7 - 1x + 14x^2 . Calculate the following values: f(a)= f(a+h)= Then simplify (f(a+h)-f(a))/h= for h does not equal 0

Question

The answer for the last part should not have an h in the denominator after you have simplified. 

 

So what is f(a)=

 

F(a+h)=

 

and f(a+h)-f(a)=

Answer 1

This process will become very familiar as you progress into calculus, as it aids in defining the meaning of a derivative.

 

First, let's calculate f(a) by substituting a in for x:

 

           f(a) = 14a² - a + 7

 

Now let's calculate f(a+h) by substituting (a+h) in for x:

 

           f(a+h) = 14(a+h)² - (a+h) + 7

 

           f(a+h) = 14(a² + 2ah + h²) - a - h + 7

 

           f(a+h) = 14a² + 28ah + 14h² - a - h + 7

 

Now f(a+h) - f(a) = (14a² + 28ah + 14h² - a - h + 7) - (14a² - a + 7)

 

Rearranging, we get f(a+h) - f(a)  = 14a² - 14a² + 28ah + 14h² - a + a - h + 7 - 7

 

                                                  = 28ah + 14h² - h

 

f(a+h) - f(a)      = 28ah + 14h² - h     = 28a + 14h - 1   For h not equal to 0

        h                            h

                              

Answer 2

 f( x ) = 14X² -X + 7

 

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

 

   [ 14 ( X +h )² - ( X + h ) + 7  - 14 X² +x - 7 ] / h =

 

  ( 14 X²  + 28 Xh + 14h²  - X - h + 7 -14x² + X - 7 ) / h =

 

    (  28 X h  + 14 h² - h ) / h =

 

        28 X -1 + 14 h

 

      You have to do a lot of the similar quadratics to be able to follow algebraic manipulation.

      In calculus you do a lot of this.

      The last expression is the limit when h approaches 0, that is 28X -1 which is the derivative of the quadratic..  

 

Answer 3

Hi Jamey,

 

The variable in the parentheses tells you what you are substituting for x in the function.

 

In the first case, f(a), you substitute x = a. 

 

So, if f(x) = 7 -x + 14x²

Then f(a) = 7 -a + 14a²  or f(a) = 14a² - a + 7 

(The convention is to write algebraic expressions in decreasing order of powers)

 

For f(a+h) we substitute x = a+h.

 

   f(x)   = 14x²          - x       + 7 

f(a+h)  = 14 (a+h)² - (a+h) + 7

           = 14(a² + 2ah + h²) - a - h + 7  

Note:  Make sure to multiply ALL terms in the parentheses by 14!

f(a+h)  = 14a² + 28ah + 14h² - a - h + 7    

 

Next we can calculate f(a+h) - f(a),

 

f(a+h) - f(a) = 14a² + 28ah + 14h² - a - h + 7 - (14a² - a + 7)

Note: Make sure to multiply ALL terms in the parentheses by -1

        

                   = 14a² + 28ah + 14h² - a - h + 7 - 14a² + a - 7

                   = 14a² + 28ah + 14h² - a - h + 7 - 14a² + a - 7

 

The 14a² and a terms cancel out and so do the 7's  to leave us with:

 

f(a+h) - f(a) = 28ah + 14h² - h  = 14h² + 28ah - h

 

(f(a+h) - f(a)) /h = 14h - 28a - 1 = 14(h - 2a) - 1

 

 

 

 

Answer 4

Reorder as f(x) = 14x^2-x+7

f(a)=14a^2-a+7

f(a+h)=14(a+h)^2-(a+h)+7

 

[f(a+h)-f(a)]/h = ([14(a+h)^2-(a+h)+7] - [14a^2-a+7])/h

= [14(a^2+2ah+h^2)-a-h+7-14a^2+a-7]/h

= [14a^2+28ah+14h^2-a-h+7-14a^2+a-7]/h

= [28ah+14h^2-h]/h

 

Divide out the 'h' to get:

= 28a+14h-1

 

Take the limit as h --> 0 to get:

= 28a-1

 

Double check using a calculus derivative, which gives:

f'(x)=28x-1

f'(a)=28a-1

They are the same.

 

DONE

Answer 5

Hi Jamey-

So when you have a function f(__), you replace whatever is in parentheses with "x" in the equation.

 

So: f(a) = 7-1a+14a^2

 

And f(a+h) = 7- 1(a+h) + 14(a+h)^2

=7-a-h + 14(a^2 + 2ah + h^2)

= 14a^2 + 28ah - a + 14h^2 - h +7

 

Unfortunately, this cannot be simplified any further.

 

Since we know f(a+h) and f(a) from above, we can take [f(a+h)-f(a)]/h

=[14a^2 + 28ah - a +14h^2 - h + 7 - (7 - a + 14a^2)]/h

 

Distribute the negative and cancel the 7s, a's, and 14^2s 

=[28ah - 14h^2 - h ]/h

=28a - 14h - 1