Find the value of [𝑓(𝑥)−𝑓(3)]/x-3 for each of the following functions

Question

Find the value of [𝑓(𝑥)−𝑓(3)]/x-3 for each of the following functions:

a) 𝑓(𝑥)=3𝑥+7 

b) 𝑓(𝑥)=3𝑥^2 −2𝑥+1

c) 𝑓(𝑥)=6/x

please show works

Answer 1

For a) we have 𝑓(𝑥)=3𝑥+7 and 𝑓(3)=3(3)+7=9+7=16. Substituting these values into the fraction we have

( 𝑓(𝑥) - 𝑓(3) ) / ( x - 3 ) = ( 3𝑥+7 - 16 ) / (x - 3 ) = ( 3x - 9 ) / ( x - 3 )

Notice that we can factor 3 from the numerator, so then  

( 3x - 9 ) / ( x - 3 ) = 3( x - 3 ) / ( x - 3 )

and now that ( x - 3 ) is clearly in both the numerator and denominator (and assuming that x is not equal to 3) they cancel and we are left with 3:

3( x - 3 ) / ( x - 3 ) = 3 [ ( x - 3 ) / ( x - 3 ) ] = 3(1) = 3

For b) we have that 𝑓(𝑥)=3𝑥^2 −2𝑥+1 and 𝑓(3) = 3(3)^2 −2(3)+1 = 3(9) - 6 + 1 = 27 - 6 + 1 = 22

We then substitute the values:

( 𝑓(𝑥) - 𝑓(3) ) / ( x - 3 ) = ( 3𝑥^2 −2𝑥+1 - 22 ) / (x - 3 ) = ( 3𝑥^2 −2𝑥 - 21 ) / (x - 3 )

Turning our attention to the numerator, if we can factor 3𝑥^2 - 2𝑥 - 21 then we may be able to simplify the entire expression. Because the last term is negative we know the factorization will have different operators like this:

( ____ + ____ )( ____ - ____ )

We know that 3x and x are the first terms of each binomial and that 3 and 7 are the last terms of each binomial, and noting that 3x(-3) + x(7) = -9x+7x = -2x we can put the terms in their proper place:

( 3x + 7)( x - 3 )

Having factored the numerator we see that ( x - 3 ) is in both the numerator and denominator, so they cancel and we arrive at our answer:

( 3𝑥^2 −2𝑥 - 21 ) / (x - 3 ) = [ ( 3x + 7)( x - 3 ) ] / ( x - 3 ) = ( 3x + 7 ) [( x - 3 ) / ( x - 3 )] = ( 3x + 7 ) [ 1 ]

= 3x + 7

Finally to c) we have that 𝑓(𝑥)=6/x and  𝑓(3)=6/(3) = 2. Substituting these values we have:

[ 𝑓(𝑥) − 𝑓(3) ] / ( x - 3 ) = [ 6/x - 2 ] / ( x - 3 )

Focusing on 6/x - 2 we multiply 2 by 1 = x/x so then

6/x - 2( x/x ) = 6/x - (2x)/x = (6 - 2x)/x

we can factor out a -2 from the numerator of this expression to obtain

(6 - 2x)/x = -2( x - 3 ) / x

Substituting this back into our expression we have:

[ 6/x - 2 ] / ( x - 3 ) = [-2( x - 3 ) / x] / ( x - 3 ) = [-2( x - 3 )] / [ x ( x - 3 ) ]

and recognizing that ( x - 3 ) is in both the numerator and denominator we let them cancel and we are left with our answer:

[-2( x - 3 )] / [ x ( x - 3 ) ] = (-2/x)[ ( x - 3 ) / ( x - 3 ) ] = (-2/x)[ 1 ] = -2/x

Answer 2

Mary:

a) f(x) = 3x + 7, f(3) = 3*3 + 7 = 16

(f(x)-f(3))/(x-3) = (3x+7-16)/(x-3) = (3x - 9)/(x-3) = 3*(x-3)/(x-3) = 3

Use same principle for other questions