Question

@Psymon Okay a little different this time :O?

Answer 1

f(x+h)-f(x)/h
Solve if f(x)=1-5x^3

Answer 2

Well, before we even start, one thing you should know. The "-f(x)" part ALWAYS cancels out. Whatever the "-f(x)" part is it better all go away or you did it wrong.

Answer 3

Okay :O I didn't know that but that makes sense now that I look at it.

Answer 4

I haven't done this in a while, so to find f(x+h), you just simply put (x+h) where x is? So f(x+h)=1-5(x+h)^3?

Answer 5

So right, otherwise is the same thing as before, plugging in (x+h) in for x and then doing -f(x)
\[\frac{ 1-5(x+h)^{3}-[1-5x ^{3}] }{ h } \]Now the cubed part is kind of annoying, but we can multiply it out the ong way or pascal's triangle with it. Either way, it should turn out like:

\[\frac{ 1-5(x ^{3}+3x ^{2}h+3xh ^{2}+h ^{3})-(1-5x ^{3}) }{ h }\]

\[\frac{ 1-5x ^{3}-15x ^{2}h-15xh ^{2}-5h ^{3}-1+5x ^{3} }{ h }\]
And as planned, we get the 1 and the 5x^3 portions to cancel out:

\[\frac{ -15x ^{2}h-15xh ^{2}-5h ^{3} }{ h }\]Factor out an h
\[\frac{ h(-15x ^{2}-15xh-5h ^{2}) }{ h }\]
\[-15x ^{2}-15xh - 5h ^{2} \]

Answer 6

Yikes! Thats still a bad equation to end with haha. One second let me write it down and see if I have questions

Answer 7

you can actually prove using factoring theorems that the end result cancels out certain parts if the expression if polynomial

Answer 8

Alrighty, lol.

Answer 9

Got it:)

Answer 10

Awesome :3

Answer 11

Okay so, next question the function is f(x)-3x^2-2x-1? How do I find f(x+h)

Answer 12

You would have to plug (x+h) into both x's in the function and then do the same thing.

Answer 13

subsitute all x's with (x+h)'s and evaluate the resulting expression

Answer 14

;_; Thats just so much work for nothing

Answer 15

unfortunately it is; in calculus you will learn a much faster way to do what you are doing - in more accurate words you learn the purpose of all of this.

Answer 16

I'm in applied calculus haha, this is just the first chapter. I haven't taken a math class in two years.

Answer 17

So f(x)=3x^2-2x-1
-3(x+h)^2-2(x+h)-1
-3(x+h)(x+h)-2x-2h-1
h^2+2hx+x^2=(x+h)^2
-3(h^2+2hx+x^2)
[-3h^2-6hx-3x^2-2x-2h-1]-3x^2-2x-1
Cancel
(-3h^2-6hx-2h)/h
-3h-6x-2

Answer 18

I didnt work it out to see what the h-term should be, but the rest of it I guarantee is right. I doubt you made a mistake considering what ya got.