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

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- myininaya

so you can find the tangent line you are given two points and the slope
use point-slope form of a line

- anonymous

Y1-Y0=m(x1-xo)?

- myininaya

actually you only need the slope

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## More answers

- anonymous

yeah

- myininaya

since that is what it asked for

- myininaya

you know the slope formula right?

- myininaya

\[m=\frac{y_1-y_2}{x_1-x_2}\]

- anonymous

m=y1-y0/x1-xo

- anonymous

is the other way around of how you put it

- myininaya

1 and 0's
1 and 2's
whatever

- myininaya

do you know how to use the formula?

- anonymous

is 1/4

- myininaya

let's see top is 2-1
and bottom is 4-0
looks great

- anonymous

yes

- anonymous

do I use y=mx+b?

- anonymous

b=2 if I do that with (4,2)

- myininaya

you 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

unless you want to find the tangent line ?

- anonymous

I do

- myininaya

it is just:
\[y-f(a)=f'(a)(x-a) \text{ is the tangent line \to the curve } y=f(x) \text{ at } (a,f(a))\]

- myininaya

you are given a is 4 here

- myininaya

\[y-f(4)=f'(4)(x-4)\]

- myininaya

now you found f(4) and f'(4)

- anonymous

right

- anonymous

thanks

- myininaya

np

- anonymous

problem the answer is not 2.

- anonymous

I already tried that

- myininaya

f(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

oh yeah then I was right about the 1/4. in the question I put the f(4)

- anonymous

perfect thank you

- myininaya

yeah 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

Find
f '(a).
f(x) =
(1 − 2x)^1/2

- myininaya

so do you know chain rule?

- anonymous

yeah, saw this last year tho

- anonymous

I thought it was 1/2(1-2x)^-1/2 * (2)

- myininaya

\[h(x)=f(g(x)) \\ h'(x)=g'(x) \cdot f'(g(x)) \\ \\ \text{ if } f(x)=x^n \\ \text{ then we have } \\ h(x)=(g(x))^n \\ h'(x)=g'(x) \cdot n(g(x))^{n-1}\]

- myininaya

well one complaint

- myininaya

what is the derivative of 1-2x ?

- myininaya

you put 2 but it should actually be...

- anonymous

ooooh

- anonymous

thanks

- myininaya

but you can also simplify a bit

- anonymous

hahaha hate when that happens

- anonymous

yes i did before

- anonymous

in my homework

- myininaya

ok then you got it now :)

- anonymous

do you have time for more?

- anonymous

I had problems with this one too

- myininaya

maybe one more then I have to go check on something

- anonymous

(a) Find the slope m of the tangent to the curve
y = 7/
x
at the point where x = a > 0.

- myininaya

is that just y=7/x?

- anonymous

(b) Find equations of the tangent lines at the points (1, 7) and
(4, 7/2)
.

- myininaya

\[y=\frac{7}{x}=7 \cdot \frac{1}{x} =7 x^{-1} \text{ by law of exponents } \\ \text{ now differentiate } y=7 x^{-1} \text{ using power rule }\]

- anonymous

no x^1/2 sorry

- myininaya

oh

- anonymous

it didn't appear when i copy pasted it

- anonymous

I had -7/2x^3/2

- anonymous

but is not that

- myininaya

\[\text{ find the slope of the tangent line \to the curve } y=7x^\frac{1}{2} \text{ at } x=a>0\]?

- myininaya

or is that x^(1/2) in the denominator?

- anonymous

ok part A i got it

- anonymous

I had to change x=a

- anonymous

now for B I don't now what I did wrong

- myininaya

ok so it was:
\[y=\frac{7}{x^\frac{1}{2}}=7x^\frac{-1}{2}\]

- anonymous

I got that

- anonymous

but I have part B wrong

- myininaya

ok the tangent line at (1,7)
is
\[y-f(a)=f'(a)(x-a) \\ \text{ here you have } a=1 \\ y-f(1)=f'(1)(x-1) \\ \text{ you already found } f'(x) \text{ \above } \\ \text{ you got that } f'(x)=\frac{-7}{2x^{\frac{3}{2}}} \text{ I think but I don't really know anymore } \\ \text{ \because I'm still trying \to figure out what } f(x) \text{ was :p } \\ \text{ so this becomes } \\ y-7=\frac{-7}{2(1)^\frac{3}{2}}(x-1)\]

- myininaya

and you still have to do this same process for the other point

- anonymous

perfect

- anonymous

is not that apparently

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