OpenStudy (anonymous):

please help me! photo attached

4 years ago
OpenStudy (anonymous):

I don't see a photo? @Dodo1

4 years ago
OpenStudy (anonymous):
4 years ago

OpenStudy (anonymous):
4 years ago

OpenStudy (anonymous):

I attached two photos, its 2 questions"

4 years ago
OpenStudy (anonymous):

@genius12

4 years ago
zepdrix (zepdrix):

\(\large 65)\) \(\large u=f(g(x))\) \(\large a)\) \(\large u'(1)\). Ok so we'll want to find the derivative of u by applying the chain rule. Remember how to do that?

4 years ago
OpenStudy (anonymous):

yes i do f'(g(x)*g(x)'

4 years ago
zepdrix (zepdrix):

Ok good. From there they want us to find \(\large u'(1)\) Which will be given by \(\large f'(\color{royalblue}{g(1)})g'(1)\). So let's try to read this piece by piece. Start with \(\large \color{royalblue}{g(1)}\). What is the value?

4 years ago
OpenStudy (anonymous):

Yes, g(1)=3?

4 years ago
zepdrix (zepdrix):

Ok good, that changes our expression to this, \(\large f'(\color{royalblue}{3})g'(1)\)

4 years ago
zepdrix (zepdrix):

Luckily these are straight lines, so even without a function we can easily determine the slope of these points. What is the slope of that line segment of \(\large f\) at \(\large x=3\) ?

4 years ago
zepdrix (zepdrix):

Just count the boxes if you're not sure. It looks like we're moving DOWN1 for every 4RIGHT yes? We can think of this as -1UP and 4RIGHT. Rise/Run = -1/4 This is our slope. Our \(\large f'(3)\). Understand how I got that value? <:o

4 years ago
OpenStudy (anonymous):

Mmmm im confsued.

4 years ago
OpenStudy (anonymous):

g(1)=1? f(1)=3.

4 years ago
OpenStudy (anonymous):

i got this part so far

4 years ago
zepdrix (zepdrix):

No, g(1)=3, as you said earlier.

4 years ago
OpenStudy (anonymous):

ops, yes.

4 years ago
OpenStudy (anonymous):

i was going to say f(1)=2

4 years ago
OpenStudy (anonymous):

First thing we notice is that each function is piece-wise and has corners so will not be differentiable at certain points. We know that u(x) = f(g(x)). Since a) asks us for u'(1), we look at functions f(x) and g(x) from (0, 2). We notice that f(x), in the interval (0, 2), is the same as saying f(x) = 2x. In the same interval (0, 2), we notice that g(x) = -3x + 6. Since we are asked for u'(1), we must first find f(g(x)) in the interval (0, 2). f(g(x)) = u(x) = -6x + 12 Therefore, u'(x) = -6 u'(1) = -6 For v'(1), we do the exact same as above. We are still using the interval (0, 2) except this time it's g(f(x)). g(f(x)) = v(x) = -3(2x) + 6 = -6x + 6 Therefore, v'(x) = -6 v'(1) = -6 For w'(1), we are doing the same thing in the interval (0, 2). This time only, we are finding g(g(x)). g(g(x)) = w(x) = -3(-3x + 6) + 6 = (9x - 18) + 6 = 9x - 12 Therefore, w'(x) = 9 w'(1) = 9 Hope that helps. @Dodo1

4 years ago
OpenStudy (anonymous):

hold on, for the first graph the answer is 3/4

4 years ago
OpenStudy (anonymous):

What graph are you talking about? I am talking about question 65.

4 years ago
OpenStudy (anonymous):

not 65, the graph of u. u=3/4 or something

4 years ago
OpenStudy (anonymous):

Which graph? The plotted in red? I was answering the other question 65. where it asks for derivatives.

4 years ago
OpenStudy (anonymous):

the pix of 1917?

4 years ago
OpenStudy (anonymous):

What's 1917 lol -.-

4 years ago
OpenStudy (anonymous):

the drivative askins u' question right? the answer is 3/4

4 years ago
OpenStudy (anonymous):

sorry asking

4 years ago
OpenStudy (anonymous):

@zepdrix any ideas?

4 years ago
OpenStudy (anonymous):

for 65, the answer for u is 3/4 but i dont know how i get to this answer

4 years ago
OpenStudy (anonymous):

The question asking u'(1) v'(1) w'(1), that's what I answered.

4 years ago
OpenStudy (anonymous):

yes, but my answer book says thst u'(1)= 3/4

4 years ago
OpenStudy (anonymous):

:/

4 years ago
OpenStudy (anonymous):

Lol.

4 years ago
OpenStudy (anonymous):

thanks for your time tho.

4 years ago
OpenStudy (anonymous):

What does it say for w'(1) and v'(1)?

4 years ago
OpenStudy (anonymous):

for v says DNE. and for w, -2

4 years ago
OpenStudy (anonymous):

this is tricky question is it?

4 years ago
OpenStudy (anonymous):

This isn't tricky. The answers just don't make any sense to me. 0,o What I got should be correct.

4 years ago
zepdrix (zepdrix):

\(\large f'(\color{royalblue}{g(1)})g'(1) \qquad = \qquad f'(\color{royalblue}{3})g'(1)\qquad = \qquad -\dfrac{1}{4}(2) \qquad = \qquad -\dfrac{1}{2}\) The book says u'(1) is 3/4??? Hmm...

4 years ago
zepdrix (zepdrix):

Oh oh oh i see, i made a stupid mistake :)

4 years ago
OpenStudy (anonymous):

yes! it does, it says use slope the slope is (2,4)(6,3) so the slope is 3-4/6-2 g is linear from (0,6)(2,0)

4 years ago
OpenStudy (anonymous):

you got it?? :)

4 years ago
zepdrix (zepdrix):

g'(1)=-3. Which gives us,\[\large -\frac{1}{4}(-3)\]

4 years ago
zepdrix (zepdrix):

I'm not really sure how to explain this to you though :( Do you remember how to find the slope of a line?

4 years ago
zepdrix (zepdrix):

Given two points*

4 years ago
OpenStudy (anonymous):

ok i will write what book says.. i dont get it at all tho

4 years ago
OpenStudy (anonymous):

to find f'(3) note that f is linear from (2,4)(6,3) so its lope is 3-4/6-2 =-1/4

4 years ago
OpenStudy (anonymous):

i have no idea why the slope is nesscary to solve this question

4 years ago
zepdrix (zepdrix):

The derivative represents the slope of the line at a given point.

4 years ago
OpenStudy (anonymous):

given point?

4 years ago
zepdrix (zepdrix):

For example, g(x)=2x from the interval [0,2] as genius mentioned. We know that the derivative will give us, g'(x)=2. This is telling us that the slope of that line segment is 2. If I were to ask you what is the value of g'(1)... I'm asking, what is the slope of the line segment at x=1. Well we determined that the entire line segment has a slope of 2 in the given interval. So at the point x=1, the slope will also be 2. A way we could verify that is by plugging 1 into our derivative. g'(1)=2.

4 years ago
OpenStudy (anonymous):

oh i see, thank you so much uguys. I have test tomorrow. wish me luck :)

4 years ago