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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

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5 Answers

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)- (a+h) + 7
 
           f(a+h) = 14(a+ 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 + 14h- 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
                              
 f( x ) = 14X-X + 7
 
   [f ( X +h) - f (x)] / ( X + h - x) =
 
   [ 14 ( X +h )2 - ( X + h ) + 7  - 14 X+x - 7 ] / h =
 
  ( 14 X + 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..  
 
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
 
 
 
 
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

Comments

I will have to use strikeout and bold in my next post.  Nice!
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