A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
If the tangent line to y = f(x) at (4, 2) passes through the point (0, 1), find f(4) and f '(4).
f(4)=2 , f'(4)=?
anonymous
 one year ago
If the tangent line to y = f(x) at (4, 2) passes through the point (0, 1), find f(4) and f '(4). f(4)=2 , f'(4)=?

This Question is Closed

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2so you can find the tangent line you are given two points and the slope use pointslope form of a line

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2actually you only need the slope

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2since that is what it asked for

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2you know the slope formula right?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[m=\frac{y_1y_2}{x_1x_2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is the other way around of how you put it

myininaya
 one year ago
Best ResponseYou've already chosen the best response.21 and 0's 1 and 2's whatever

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2do you know how to use the formula?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2let's see top is 21 and bottom is 40 looks great

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0b=2 if I do that with (4,2)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2you don't need to it just asked to find f'(4) and f(4) and f'(4) is the slope of the tangent line at (4,2) to the curve y=f(x)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2unless you want to find the tangent line ?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2it is just: \[yf(a)=f'(a)(xa) \text{ is the tangent line \to the curve } y=f(x) \text{ at } (a,f(a))\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2you are given a is 4 here

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[yf(4)=f'(4)(x4)\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2now you found f(4) and f'(4)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0problem the answer is not 2.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I already tried that

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2f(4) should be 2 it is given by (4,2) f'(4) should be 1/4 that is the slope of the tangent line at (4,2) to the curve y=f(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh yeah then I was right about the 1/4. in the question I put the f(4)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2yeah you already found f(4) earlier we were finding f'(4) and we used the slope formula because that is what the derivative means

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find f '(a). f(x) = (1 − 2x)^1/2

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2so do you know chain rule?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, saw this last year tho

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thought it was 1/2(12x)^1/2 * (2)

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[h(x)=f(g(x)) \\ h'(x)=g'(x) \cdot f'(g(x)) \\ \\ \text{ if } f(x)=x^n \\ \text{ then we have } \\ h(x)=(g(x))^n \\ h'(x)=g'(x) \cdot n(g(x))^{n1}\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2what is the derivative of 12x ?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2you put 2 but it should actually be...

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2but you can also simplify a bit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hahaha hate when that happens

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2ok then you got it now :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you have time for more?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I had problems with this one too

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2maybe one more then I have to go check on something

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(a) Find the slope m of the tangent to the curve y = 7/ x at the point where x = a > 0.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(b) Find equations of the tangent lines at the points (1, 7) and (4, 7/2) .

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[y=\frac{7}{x}=7 \cdot \frac{1}{x} =7 x^{1} \text{ by law of exponents } \\ \text{ now differentiate } y=7 x^{1} \text{ using power rule }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it didn't appear when i copy pasted it

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ find the slope of the tangent line \to the curve } y=7x^\frac{1}{2} \text{ at } x=a>0\]?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2or is that x^(1/2) in the denominator?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now for B I don't now what I did wrong

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2ok so it was: \[y=\frac{7}{x^\frac{1}{2}}=7x^\frac{1}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but I have part B wrong

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2ok the tangent line at (1,7) is \[yf(a)=f'(a)(xa) \\ \text{ here you have } a=1 \\ yf(1)=f'(1)(x1) \\ \text{ you already found } f'(x) \text{ \above } \\ \text{ you got that } f'(x)=\frac{7}{2x^{\frac{3}{2}}} \text{ I think but I don't really know anymore } \\ \text{ \because I'm still trying \to figure out what } f(x) \text{ was :p } \\ \text{ so this becomes } \\ y7=\frac{7}{2(1)^\frac{3}{2}}(x1)\]

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2and you still have to do this same process for the other point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is not that apparently
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.