anonymous
  • anonymous
???
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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freckles
  • freckles
you can find the tangent line at x=e the only problem is when differentiating f(x)=2ln(x) and f'(e) is probably something you would want to use calculator for but I was told the whole point of these exercises was to approximate without using calculator I guess we could just use the approximate for e as 2.72 when it comes to that anyways the tangent line at x=e will look something like this: y-f(e)=f'(e)(x-e) y=f(e)+f'(e)(x-e) so f(x) is approximately f(e)+f'(e)(x-e)
anonymous
  • anonymous
So I would plug in values into that?
freckles
  • freckles
you can use that approximation equation to approximate f(2) f(2) is approximately f(e)+f'(e)(2-e)

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freckles
  • freckles
you still need to find f(e) and f'(e) though
anonymous
  • anonymous
How would I find those. And would it help if i showed the answer choices?
freckles
  • freckles
there are choices for approximating f(2)?
freckles
  • freckles
f(e) can bound by pluggin in e into f(x)=2ln(x) f'(x) needs to be found before finding f'(e)
anonymous
  • anonymous
y = x − 2 y = x − 2 + ln4 y = 2ln(x − 2) y = 1 This is what it shows
freckles
  • freckles
oh they want the linear approximation for f(2)
freckles
  • freckles
they want the tangent line at x=2
freckles
  • freckles
so just find the tangent line at x=2
anonymous
  • anonymous
When I derive 2lnx I get 2/x
freckles
  • freckles
right now plug in 2
freckles
  • freckles
\[y-f(2)=f'(2)(x-2) \\ y=f(2)+f'(2)(x-2)\]
anonymous
  • anonymous
I got 1 when I plugged it in
freckles
  • freckles
ok so f'(2)=1 you still need to find f(2)
freckles
  • freckles
\[y=f(2)+1(x-2) \\ y=f(2)+(x-2) \\ y=f(2)+x-2 \\ y=x-2+f(2)\]
freckles
  • freckles
f(x)=2ln(x) input 2 to find f(2)
anonymous
  • anonymous
1.39
freckles
  • freckles
don't approximate plug in 2
freckles
  • freckles
then just use power rule for log
freckles
  • freckles
\[a \ln(x)=\ln(x^a) \text{ power rule }\]
freckles
  • freckles
\[f(2)=2\ln(2)=...\]
anonymous
  • anonymous
When I plugged that into my calculator I got 1.39
freckles
  • freckles
yes that is great but we are not approximating use power rule bring the 2 up as an exponent of what is inside the log
freckles
  • freckles
example \[5\ln(2)=\ln(2^5)=\ln(32)\]
freckles
  • freckles
you try 2ln(2)=what using power rule
anonymous
  • anonymous
So I would do ln(2^2)=ln(4)
freckles
  • freckles
yes
freckles
  • freckles
the tangent line at x=a looks like this y-f(a)=f'(a)(x-a) so we has 2 instead of a y-f(2)=f'(2)(x-2) you found f(2) as ln(4) and f'(2) as 1 y-ln(4)=1(x-2) y-ln(4)=x-2 add ln(4) on both sides and you should see clearly see your answer
anonymous
  • anonymous
x-2+ln4?
freckles
  • freckles
yes
freckles
  • freckles
y=x-2+ln(4)
anonymous
  • anonymous
Thank you so much!!!!
freckles
  • freckles
np

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