f(x)=2x^2+x if you plug in (x+h) for all the X's will you get 3x^2+3h^2+4xh
there is no 3 in it
\[f(x)=2x^2+x\] \[f(x+h)=2(x+h)^2+(x+h)\]
Right but when i did that i distributed first (x+h)(x+h) and got x^2+xh+xh+h^2
when you multiply out you get \[2(x+h)(x+h)+x+h=2(x^2+2xh+h^2)+x+h\] \[=2x^2+4xh+2h^2+x+h\] there are not like terms to combine here
i think i see what you did you added \(2h^2+h\) and got \(3h^2\) but they are not like terms, because one is a square and the other isn't so you cannot add them in to one term
oh okay okay so after that i have to (2x2+4xh+2h2+x+h)-(2x^2+2) over h I have no clue how to complete it
now you have like terms
i assume it is actually \[\frac{f(x+h)-f(x)}{h}\] am i right?
yes
then you had a typo there you wrote \[(2x^2+4xh+2h2+x+h)-(2x^2+2)\] but it should be \[(2x2+4xh+2h2+x+h)-(2x^2+x)\]
i.e. the \(x\) instead of a \(2\) at the end
oh yea Once i simplify i gotten 4xh+2h^2+h correct?
\[(2x2+4xh+2h2+x+h)-(2x^2+x)\]distribute the minus sign and get \[2x^2+4xh+2h2+x+h-2x^2-x\]
you are right, yes
\[4xh+2h^2+h\] is correct for the numerator yes
so how am i suppose to divide 4xh/h
cancel an \(h\) from each term, or factor out the common factor of \(h\) in the numerator and cancel all at once
\[\frac{4xh}{h}=\frac{4x\cancel h}{\cancel h}=4x\] for example
or you can write \[\frac{4xh+2h^2+h}{h}=\frac{h(4x+2h+1)}{h}\] then cancel looks better that way probablyl
so i should end up with 4x+h^2 right
no
is this step clear? factoring out an \(h\) in the numerator?\[\frac{4xh+2h^2+h}{h}=\frac{h(4x+2h+1)}{h}\]
okay i dont under stand so after i factor it out then what
cancel the \(h\)
\[\frac{4xh+2h^2+h}{h}=\frac{h(4x+2h+1)}{h}=\frac{\cancel h(4x+2h+1)}{\cancel h}=4x+2h+1\]
thank you soooooooo much!
Join our real-time social learning platform and learn together with your friends!