Question

Can someone explain to me how to do this problem:

Find an equation of the line that is tangent to the graph of f and parallel to the given line.

f(x) = 2x^2 4x+y+3=0

Answer 1

Lets work on this question together

Answer 2

Now to find the tangent line to the graph of f we need to find the derivative of f. For this particular function you can the power rule. Can you find the derivative?

Answer 3

\[f \prime (x) = 4x\]

Answer 4

correct

Answer 5

now, since we are looking for tangent line that is parallel to the given line, we have to find the slope of that given line, can you find the slope of the given line?

Answer 6

4x

Answer 7

are you sure?? check again

Answer 8

?

Answer 9

4

Answer 10

Well, thats still not quite right: here it is,
you have 4x+y+3=0, now lets solve for y to get this into y=mx+b form.
Subtract 4x from both sides and you will get: y+3=-4x.
Now subtract 3 from both sides and you will get: y=-4x-3. Thus our slope is -4.

Answer 11

Now since the slope of this is -4, we can set it equal to f'(x) because parrallel lines mean that the slope is equal.

Answer 12

ok

Answer 13

so, we have :
4x=-4

Answer 14

can you solve for x?

Answer 15

-1

Answer 16

correct

Answer 17

now, lets find the y coordinate, by plugging negative 1 back into the orginal equation: can you evluate f(-1): 2x^2

Answer 18

-2x^2

Answer 19

well, remeber f(-1) means that you replace all the x with -1, So we have: f(-1)=2(-1)^2.
Can you evluate that?

Answer 20

4

Answer 21

oh yeah, I forgot about that. In that case its 4.

Answer 22

order of operations: exponents first, then multiplication

Answer 23

Thus: 2*(-1)^2

Answer 24

square -1 first, then multiply by 2

Answer 25

2 then

Answer 26

correct

Answer 27

now we have what we need, we have the y-coordinate: 2
x:coordinate: -1
and the slope: -4

Answer 28

now with this information we can use the point slope form to get the equation of the tangent line to f, can you do that?

Answer 29

\[2 = -4(1)+b\]\[b=6, y = -4x+6\]

Answer 30

y-2=-4(x-(-1))

Answer 31

y-2=-4(x+1)

Answer 32

y-2=-4x-4

Answer 33

oh, so I have to do it that way then. Okay...

Answer 34

y=-4x-2

Answer 35

and that is your final answer

Answer 36

Yes, I'm trying to find the derivative of this function using the chain rule and the product rule. \[y = x \sqrt{1-x^{2}}\]

Answer 37

okay then, so the product rule goes like F'*S+S'*F.
F=first term
S=second term

Answer 38

so lets do this: what is the derivative of the first term x?

Answer 39

1

Answer 40

right, now what is the derivative of the second term, you will have to use the chain rule here. Hint: write the second term as (1-x^2)^1/2

Answer 41

\[1/2(1-x)^{-1/2}(-2x)\]

Answer 42

it should be 1-x^2 inside the first parentheses right?

Answer 43

should be: \[1/2(1-x^2)^-1/2*(-2x)\]

Answer 44

4

Answer 45

yes, sorry bout that

Answer 46

okay, now lets finish it up by using the product rule

Answer 47

we have : (1)(1-x^2)^1/2+ 1/2(1-x^2)^-1/2(-2x)(x)

Answer 48

we can clean this up: (1-x^2)^1/2+(-2x^2)./2(1-x^2)^1/2

Answer 49

now all we do is get lcd. Can you do that?

Answer 50

Does that last part say 2(1-x^2)^1/2?

Answer 51

yes

Answer 52

So does that whole thing say: \[(1-x ^{2})^{1/2}+(-2x ^{2})/2(1-x ^{2})^{1/2}\]

if so how did you get \[2(1-x ^{2})^{1/2}\]

Answer 53

yes thats righ

Answer 54

look at the post right underneath where i write"okay, now lets finish it up by using the product rule"

Answer 55

basically, first i factores out the 1/2, then i just mulitplied (-2x)*(x). Next, i simply made (1-x^2)^-1/2 into (1)/(1-x^2)^1/2. Finally i multipled that by (-2x^2)/2

Answer 56

(-2x^2)/2 comes from (1/2)*(-2x^2)

Answer 57

okay I think I get it now.

Answer 58

alright:)

Answer 59

So -2x^2/2 is the answer?

Answer 60

rmeber you need to get lcd to fully simplify

Answer 61

so not the answer not -2x^2/2

Answer 62

here is what we have:
\[(1-x^2)^{1/2}-((2x^2)/(2(1-x^2)^{1/2})\]

Answer 63

now you have to find the lcd and simplify

Answer 64

so (1-x^2) is the lcd...

Answer 65

2(1-x^2)^1/2 is the lcd

Answer 66

Is this right so far: \[2(1-x ^{2})^{1/2}/2(1-x ^{2})^{1/2} \times (1-x ^{2})^{1/2}/2(1-x ^{2})^{1/2} - 2x ^{2}/2(1-x ^{2)^{1/2}}\]

Everything on the left side cancels out except for the 2 So your left with 1/2. If you flip that and bring the 2 on top you have:
\[2 \times (1) - 2x ^{2} /2(1-x ^{2})^{1/2}\]

Answer 67

the 2;s in the numerator and denominator cancel out so you are left with on (1-2x^2)/(1-x^2)^1/2

Answer 68

Then if I change that denominator to \[\sqrt{1-x ^{2}}\]

my final answer, is then \[1-2x ^{2}/\sqrt{1-x ^{2}}\]

Answer 69

that is correct