Math Problem: (Don't know how to solve this. Please explain when showing how to solve it.)

- anonymous

Math Problem: (Don't know how to solve this. Please explain when showing how to solve it.)

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- anonymous

Evaluate\[(f(x+h)-f(x))/h\] and simplify if \[f(x) = x^2-2x\]

- anonymous

this stuff is differentiation

- anonymous

i forgot how to do it this method...but i normally use power rule
\[\frac{ d }{ dx }x = x ^{n-1}\]

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## More answers

- Nnesha

hmm first you need to find f(x+h)
so substitute x for (x+h) in f(x) function \[\huge\rm f(x+h)=\color{reD}{x}^2-2\color{ReD}{x}\]

- anonymous

so i need to solve for x using that equation u gave me nnesha?

- anonymous

and i dont know that power rule saseal

- Nnesha

not for x simplify that
that's the equation i gave you
you already have the formula

- Nnesha

and i don't know that either :(

- Nnesha

\[\huge\rm f(x+h)=\color{reD}{(x+h)}^2-2\color{ReD}{(x+h)}\]
solve that use foil method for (x+h)^2 and distribute parentheses by -2

- anonymous

ok...power rule is like (ill just show for x^2)
\[\frac{ d }{ dx }(x^2) = 2x\]

- anonymous

How did you get 2x? what was the value of d @saseal?

- anonymous

actually they call it dy/dx it means to differentiate

- anonymous

So the answer for that problem would be\[(x ^{2} +2xh +h ^{2})-2x-2h\] @Nnesha?

- anonymous

so are you suppose to plug in y and x instead of d @saseal?

- Nnesha

no not the answer that's just f(x+h)
now use that formula \[\huge\rm \frac{ f(x+h)-f(x) }{ h }\]
replace f(x+h) by (x^2 +2xh+h^2)-2x-2h
and f(x) by the function x^2-2x

- Nnesha

\[\huge\rm \frac{ \color{ReD}{f(x+h)}-\color{blue}{f(x)} }{ h }\]
\[\huge\rm \frac{ \color{reD}{ x^2 +2xh+h^2-2x-2h } -\color{blue}{(x^2 -2x) }}{ h }\]
simplify !

- anonymous

\[f'(x)=\frac{((x+h)^2-2x+h)- (x^2-2x)}{ h}\]

- anonymous

(2xh+h^2-2h)/h

- anonymous

That's what I got after putting like terms together

- anonymous

ill show you the power rule magic after you get everything together

- Nnesha

there is a common factor h^2+2xh+h

- anonymous

Alright

- anonymous

Oh ya. h(h+2x+1)

- anonymous

Is that the final answer?

- anonymous

no

- anonymous

Am I suppose to factor or insert it into another equation?

- Nnesha

\[\large\rm \frac{ \color{reD}{ \cancel{x^2} +2xh+h^2\cancel{-2x}-2h } -\color{blue}{\cancel{x^2} +\cancel{2x }}}{ h }\]
\[\large\rm \frac{ h^2+2xh+h }{ h}\] \[\large\rm \frac{ h(h+2x+1)}{ h}\]

- Nnesha

can you cancel out h?

- anonymous

Oh whoops I forgot about the h on the bottom. So the answer would actually be h+2x+1?

- Nnesha

yep

- anonymous

Alright thanks nnesha and saseal!

- anonymous

where did the coefficient of -2h go?

- Nnesha

hmm let's see.

- anonymous

I think we wrote that step wrong.

- anonymous

Alright its 2x-2+h

- Nnesha

\[\large\rm \frac{ \color{reD}{ \cancel{x^2} +2xh+h^2\cancel{-2x}-2h } -\color{blue}{\cancel{x^2} +\cancel{2x }}}{ h }\]
\[\large\rm \frac{ h^2+2xh\color{ReD}{-2h} }{ h}\] \[\large\rm \frac{ h(h+2x\color{reD}{-2})}{ h}\]
i forgot it suppose to be -2h not just h

- Nnesha

thanks saseal

- Nnesha

\[\large\rm \frac{ \cancel{h}(h+2x\color{reD}{-2})}{\cancel{ h}}\]

- anonymous

Ok, as promised the power rule: you take the power of the variable and pull it down and minus 1 to the power like this
\[(\frac{ dy }{ dx })2x^2 = (2)2x ^{2-1}\]

- anonymous

Okay

- anonymous

if its just 2x it becomes
\[\frac{ dy }{ dx }(2x ^{1}) = 2x^0=2(1)\]

- anonymous

why did it become 2x^0?

- anonymous

so for your equation x^2-2x
\[\frac{ dy }{ dx }(x^2-2x) = 2x-2\]

- anonymous

because x^1 is x, after you differentiate it you need to -1 from its power so it becomes x^0

- anonymous

Okay I think I understand what you did. Would we learn this in AP Calculus AB?

- anonymous

if you wonder where the h went...its removed by limits as h approaches 0

- anonymous

Yea you would

- anonymous

it's like the meat of the calculus

- anonymous

Oh okay. Did you take AP Calculus?

- anonymous

@saseal

- anonymous

No, AP is not available in my country, we do A levels instead which is about the same

- anonymous

i don't live in us

- anonymous

Oh okay sorry about that.

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