me again

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denise

New member
Joined
Jul 2, 2005
Messages
12
i'm feelin like sooo dumb. i have several problems in which i got stuck on.

if f(x) = x^(2) + 3x - 2, then find the following:

f(x+h)

i pluged in the numbers and got

(x+h)^ (2) + 3(x + h) - 2

now i'm stuck. i tried doin other things but it didnt work out. where do i go from here.

also

f(x+h) - f(x) / x

i pluged in the numbers for this one also and i'm stuck now

(x+h)^(2) + 3(x + h) - 2 - x^(2) + 3x - 2 / x

would i factor x ^(2) + 3x -2 first or what?
 
i did that and i got

x ^(2) + +2xh + h^(2) + 3(x+h) - 2
x ^(2) + +2xh + h^(2) + 3x + 3h - 2

now what do i do?
 
x ^(2) + +2xh + h^(2) + 3x + 3h - 2

Simplify - Combine like terms - factor

x^2 + x*(2h+3) + (h^2 + 3h - 2)


On the rest:

(x+h)^(2) + 3(x + h) - 2 - x^(2) + 3x - 2 / x

This is not good for two reasons, both clarity.

It should be:

[(x+h)^(2) + 3(x + h) - 2 - (x^(2) + 3x - 2)] / x

Ar you sure that is divided by 'x', or was it 'h'?
 
The way you show it: f(x+h) - f(x) / x
means: f(x+h) - [f(x) / x]
so:
(x+h)^2 + 3(x+h) - 2 - [(x^2 + 3x - 2) / x]

x^2 + 2hx + h^2 + 3x + 3h - 2 - [(x^2 + 3x - 2) / x]

[x^3 + 2hx^2 + xh^2 + 3x^2 + 3hx - 2x - x^2 - 3x + 2] / x

[x^3 + 2hx^2 + 2x^2 + xh^2 + 3hx - 5x + 2] / x

[x^3 + 2x^2(h + 1) + x(h^2 + 3h - 5) + 2] / x

IF BOTH terms are divided by x, then this is the way
(as TK told you) to show: [f(x+h) - f(x)] / x

...and that's much easier, since most terms cancel out leaving:
h^2 + 2hx + 3h = h(h + 2x + 3)
 
x^2 + 2hx + h^2 + 3x + 3h - 2 - [(x^2 + 3x - 2) / x]

[x^3 + 2hx^2 + xh^2 + 3x^2 + 3hx - 2x - x^2 - 3x + 2] / x


how did u get this from the equation above?
 
denise said:
x^2 + 2hx + h^2 + 3x + 3h - 2 - [(x^2 + 3x - 2) / x]
[x^3 + 2hx^2 + xh^2 + 3x^2 + 3hx - 2x - x^2 - 3x + 2] / x
how did u get this from the equation above?
Click to expand...

the same way as:
x^2 - 2x - 4/x
= (x^3 - 2x^2 - 4) / x

or 4 - 1/2
= (8 - 1) / 2
 
x^2 + 2hx + h^2 + 3x + 3h - 2 - (x^2 + 3x - 2) / x

Multiply all those things in front by x / x, also known as 1.

x * (x^2 + 2hx + h^2 + 3x + 3h - 2) / x - (x^2 + 3x - 2) / x

Distribute the x on the left and the negative on the right.

(x^3 + 2hx^2 + xh^2 + 3x^2 + 3hx - 2x) / x + (-x^2 - 3x + 2) / x

Since the fractions have the same denominator, add the numerators.

(x^3 + 2hx^2 + xh^2 + 3x^2 + 3hx - 2x - x^2 - 3x + 2) / x
 
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