Search 73,942 tutors
0 0

# 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

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)=

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

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