anonymous
  • anonymous
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)=?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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myininaya
  • myininaya
so you can find the tangent line you are given two points and the slope use point-slope form of a line
anonymous
  • anonymous
Y1-Y0=m(x1-xo)?
myininaya
  • myininaya
actually you only need the slope

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anonymous
  • anonymous
yeah
myininaya
  • myininaya
since that is what it asked for
myininaya
  • myininaya
you know the slope formula right?
myininaya
  • myininaya
\[m=\frac{y_1-y_2}{x_1-x_2}\]
anonymous
  • anonymous
m=y1-y0/x1-xo
anonymous
  • anonymous
is the other way around of how you put it
myininaya
  • myininaya
1 and 0's 1 and 2's whatever
myininaya
  • myininaya
do you know how to use the formula?
anonymous
  • anonymous
is 1/4
myininaya
  • myininaya
let's see top is 2-1 and bottom is 4-0 looks great
anonymous
  • anonymous
yes
anonymous
  • anonymous
do I use y=mx+b?
anonymous
  • anonymous
b=2 if I do that with (4,2)
myininaya
  • 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
  • myininaya
unless you want to find the tangent line ?
anonymous
  • anonymous
I do
myininaya
  • 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
  • myininaya
you are given a is 4 here
myininaya
  • myininaya
\[y-f(4)=f'(4)(x-4)\]
myininaya
  • myininaya
now you found f(4) and f'(4)
anonymous
  • anonymous
right
anonymous
  • anonymous
thanks
myininaya
  • myininaya
np
anonymous
  • anonymous
problem the answer is not 2.
anonymous
  • anonymous
I already tried that
myininaya
  • 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
  • anonymous
oh yeah then I was right about the 1/4. in the question I put the f(4)
anonymous
  • anonymous
perfect thank you
myininaya
  • 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
  • anonymous
Find f '(a). f(x) = (1 − 2x)^1/2
myininaya
  • myininaya
so do you know chain rule?
anonymous
  • anonymous
yeah, saw this last year tho
anonymous
  • anonymous
I thought it was 1/2(1-2x)^-1/2 * (2)
myininaya
  • 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
  • myininaya
well one complaint
myininaya
  • myininaya
what is the derivative of 1-2x ?
myininaya
  • myininaya
you put 2 but it should actually be...
anonymous
  • anonymous
ooooh
anonymous
  • anonymous
thanks
myininaya
  • myininaya
but you can also simplify a bit
anonymous
  • anonymous
hahaha hate when that happens
anonymous
  • anonymous
yes i did before
anonymous
  • anonymous
in my homework
myininaya
  • myininaya
ok then you got it now :)
anonymous
  • anonymous
do you have time for more?
anonymous
  • anonymous
I had problems with this one too
myininaya
  • myininaya
maybe one more then I have to go check on something
anonymous
  • anonymous
(a) Find the slope m of the tangent to the curve y = 7/ x at the point where x = a > 0.
myininaya
  • myininaya
is that just y=7/x?
anonymous
  • anonymous
(b) Find equations of the tangent lines at the points (1, 7) and (4, 7/2) .
myininaya
  • 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
  • anonymous
no x^1/2 sorry
myininaya
  • myininaya
oh
anonymous
  • anonymous
it didn't appear when i copy pasted it
anonymous
  • anonymous
I had -7/2x^3/2
anonymous
  • anonymous
but is not that
myininaya
  • myininaya
\[\text{ find the slope of the tangent line \to the curve } y=7x^\frac{1}{2} \text{ at } x=a>0\]?
myininaya
  • myininaya
or is that x^(1/2) in the denominator?
anonymous
  • anonymous
ok part A i got it
anonymous
  • anonymous
I had to change x=a
anonymous
  • anonymous
now for B I don't now what I did wrong
myininaya
  • myininaya
ok so it was: \[y=\frac{7}{x^\frac{1}{2}}=7x^\frac{-1}{2}\]
anonymous
  • anonymous
I got that
anonymous
  • anonymous
but I have part B wrong
myininaya
  • 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
  • myininaya
and you still have to do this same process for the other point
anonymous
  • anonymous
perfect
anonymous
  • anonymous
is not that apparently

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