Question

I need help with a limit defenition problem. Will fan and medal!

I did it in derivative form and my teacher wants it the other way but I can't figure it out.

F(x)=2x^-6x+2 and x=2

I have it set up as ((2(2+h)^2-6(2)+2) - (2(2^2)+2)))/h

And don't know where to go from here.

Thank you!!

Answer 1

def of derivative is
\[\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\]

Answer 2

I know that much

Answer 3

whats the other form that teacher wants?

Answer 4

I'm sorry I wasn't clear! She wants us to use that to find the equation of the tanget line

Sorry!

Answer 5

oh ok

Answer 6

equation of tangent line:
\[ y - f(x_0) = f'(x_0) ( x - x_0)\]
x_0 = 2

Answer 7

also i think you are missing some terms in your setup
\[\rightarrow \frac{ (2(h+2)^2 - 6(h+2) +2)) - (2*2^2 - 6(2) +2)}{h}\]

Answer 8

you have to simplify this by distributing and combining like terms

Answer 9

So do I need to add the h in with the 6?

Answer 10

yes .... everywhere there is an "x" , you must replace with "h+2"

Answer 11

Would the 2 be before the H or does it matter?

Answer 12

doesn't matter ..... a+b = b+a

Answer 13

\[((12h+4^2-6h+12+2)-(16-12+2))/h\]

Answer 14

Only 2h

Answer 15

ok you need to square the (h+2) first before distributing the 2
\[(h+2)^2 = (h+2)(h+2) = h^2 +2h+2h +4\]

Answer 16

Sorry my computer died!

Answer 17

\[(4+2h)(4+h)-12+6h+2)-(16-12+2)\]


Would this be right?

Answer 18

no , remember to distribute correctly
--> -6(h+2) = -6h - 12 , the negative is multiplied by both terms

2(h+2)^2 ----> square first, then distribute the 2 (look at my last post)

Answer 19

2(2^2) = 2(4) = 8 ..... not 16

Answer 20

Oh now I see

Answer 21

\[8+2h-12+6h)+2)-8-12+2\]

Answer 22

how do you get 8 +2h ?

Answer 23

\[2(h+2)^2 = 2(h^2 +4h+4) = 2h^2 +8h+8\]

Answer 24

it's 2(2+h)^2 so 2 squared is four. four times 2 is 8.

Answer 25

I think I'm just stressed right now and not thinking straight

Answer 26

yes that is correct for the constant term.... but you have to include the variable "h" when you square it

remember FOIL, multiplying binomials (x+2)(x+2)

Answer 27

anyway you should have
\[\frac{(2h^2 +8h+8 -6h-12+2) - (8-12+2)}{h}\]
\[= \frac{2h^2+2h}{h}\]
\[= \frac{h(2h+2)}{h}\]
\[=2h +2\]
Take limit as h->0 (plug in h=0)
\[f'(2) = 2\]

Answer 28

So what would the intercept be?

Answer 29

\[y-(-2)=2(x-2)\]