Question

I need help with using limit definition to find the tangent line. 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

And don't know where to go from here.

Thank you!!

((2h2+8h+8−12+6h+2)−8−12+2)/h


this is what I have so far

Answer 1

derivative form?

Answer 2

do you mean by using the definition of derivative?

Answer 3

That's just what she called it. She said I did it in the wrong form

Answer 4

I think you mean by definition of derivative
derivative form means nothing to me

Answer 5

Also I think there is a type-o in your function

Answer 6

\[f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\]
this is definition of derivative

Answer 7

Yes

Answer 8

and she probably means she wants you to find the derivative using this definition

Answer 9

Yes

Answer 10

\[f(x)=2x^{-6x}+2\]
is this really the correct function?

Answer 11

\[2x^2-6+2\]

Answer 12

that is weird
why wouldn't they just combine like terms

Answer 13

\[2x^2-6x+2\]

Answer 14

\[f(x)=2x^2-6x+2 \\ f(x+h)=2(x+h)^2-6(x+h)+2\]
just should have been the first thing stated

Answer 15

\[\frac{f(x+h)-f(x)}{h} \\ \frac{[2(x+h)^2-6(x+h)+2]-[2x^2-6x+2]}{h} \\ \text{ you can do some multiplying out and canceling of terms in numerator first } \\ \text{ I sometimes like to factor things by rearranging the right terms together }\]

Answer 16

So combine the x's?

Answer 17

if you choose factoring way.. combine terms like this first:
\[\frac{2[(x+h)^2-x^2]-6[(x+h)-x]+[2-2]}{h}\]

if you choose to multiply things out... you should have this next:
\[\frac{2x^2+4xh+2h^2-6x-6h+2-2x^2+6x-2}{h} \]

Answer 18

and then for the multiplying things out part you collect like terms in the numerator

Answer 19

for the factoring way if you choose it you will need to know how to factor a difference of squares

Answer 20

I am still really confused on this

Answer 21

on multiplying part?

Answer 22

On all of it and how things plug in

Answer 23

\[f(x+h)=2(x+h)^2-6(x+h)+2 \\ f(x+h)=2(x+h)(x+h)-6(x+h)+2\]
do you know know to expand (x+h)(x+h) through multiplication
and -6(x+h) through multiplication ?

Answer 24

oh you aren't sure how to find f(x+h)?

Answer 25

I think it's the equation but I'm not sure

Answer 26

\[f(x)=2x^2-6x+2\]
notice the x's..
the x acts as the input value until we can think of another input value to replace it with

Answer 27

\[(2h^2+8h+8-12+6h+2)-8-12+2)/h\]

Answer 28

\[f(x)=2x^2-6x+2 \\ \text{ examples of how the machine's outputs change base on the input } \\ f(1)=2(1)^2-6(1)+2 \\ f(1+2)=2(1+2)^2-6(1+2)+2 \\ f(3)=2(3)^2-6(3)+2 \\ f(\star)=2(\star)^2-6(\star)+2 \\ f(\star+\Delta)=2(\star+\Delta)^2-6(\star+\Delta)+2 \\ f(x+h)=2(x+h)^2-6(x+h)+2\]
we are finding f(x+h) since in the definition of the derivative it has f(x+h)
...
\[f'(x)=\lim_{h \rightarrow 0} \frac{\color{red}{f(x+h)}-f(x)}{h}\]

Answer 29

\[\text{ everywhere there is an } x \text{ in } f(x)=2x^2-6x+2 \\ \text{ we replace it with } (x+h) \text{ since we are to plug in } f(x+h)\]

Answer 30

I get that part

Answer 31

\[f(x+h)=2(x+h)^2-6(x+h)+2\]
can you expand this (perform the multiplication)

Answer 32

2x+2h+2x+2h-6x+6h+2

Answer 33

how did you get that?

Answer 34

Just multiplying it out

Answer 35

how did you multiply 2(x+h)^2 out?

Answer 36

2(x+h)(x+h)

Then just kinda foiled it

Answer 37

but your x^2, h^2, and 2xh term disappeared?
2(x+h)(x+h)
2(x^2+2xh+h^2)

Answer 38

oh

Answer 39

do you understand how to multiply binomials?

Answer 40

\[(a+b)(x+y) \\ a(x+y)+b(x+y) \\ ax+ay+bx+by \\ \]

Answer 41

yes

Answer 42

\[(x+h)(x+h) \\ x(x+h)+h(x+h) \\ xx+xh+hx+hh \\ x^2+2xh+h^2\]

Answer 43

yes

Answer 44

\[f'(x)=\lim_{h \rightarrow 0} \frac{[f(x+h)]-[f(x)]}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{[2(x^2+2xh+h^2)-6x-6h+2]-[2x^2-6x+2]}{h}\]
do distributive property and then combine like terms

Answer 45

\[(2x^2+4xh+2h^2)-6x-6h+2)-2x^2-6x+2\]

Answer 46

the whole reason I put [ ] around the 2x^2-6x+2 was so you wouldn't forget that the whole f(x) is being subtract not just part of

Answer 47

\[f'(x)=\lim_{h \rightarrow 0} \frac{[f(x+h)]-[f(x)]}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{[2(x^2+2xh+h^2)-6x-6h+2]-[2x^2-6x+2]}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{2x^2+4xh+2h^2-6x-6h+2-2x^2+6x-2}{h}\]

Answer 48

combine like terms in numerator

Answer 49

\[(4xh+6h)/h\]

Answer 50

missing one more term

Answer 51

and also there is a - in front of 6h

Answer 52

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

Answer 53

\[f'(x)=\lim_{h \rightarrow 0} \frac{[f(x+h)]-[f(x)]}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{[2(x^2+2xh+h^2)-6x-6h+2]-[2x^2-6x+2]}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{\color{Red}{2x^2}+4xh+2h^2-\color{blue}{6x}-6h+\color{green}{2}-\color{red}{2x^2}+\color{blue}{6x}-\color{green}{2}}{h} \\ \]
correct

Answer 54

now you should see a common factor from numerator and denominator

Answer 55

h/h=1

Answer 56

\[\frac{h}{h} \cdot \frac{4x+4h-6}{1}\]

Answer 57

Doesn't h=0 though?

Answer 58

we will plug in h=0 once we get rid of the thing h on bottom

Answer 59

0/0 is indeterminate
this is why getting rid of that h in the denominator will give us a form that is non-indeterminate

Answer 60

4x-6?

Answer 61

4x+4-6

Answer 62

are you understanding we have 4x-6?

Answer 63

I was canceling out the h

Answer 64

So that gives us 6?

Answer 65

Is it just y=4x-6?

Answer 66

@myininaya

Answer 67

\[\frac{4xh+2h^2-6h}{h}=\frac{h}{h} \cdot \frac{4x+2h-6}{1}=(1) \cdot (4x+2h-6) =4x-6 \text{ since } h \rightarrow 0\]

Answer 68

so yes this is how I got f'(x) above

Answer 69

f'(x)=4x-6

Answer 70

f'(x)=4x-6 is the slope formula

Answer 71

function

Answer 72

so if we want the slope at x=5 evaluate f'(5)

Answer 73

Thank you