???

- anonymous

???

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

So I would plug in values into that?

- freckles

you can use that approximation equation to approximate f(2)
f(2) is approximately f(e)+f'(e)(2-e)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- freckles

you still need to find f(e) and f'(e) though

- anonymous

How would I find those. And would it help if i showed the answer choices?

- freckles

there are choices for approximating f(2)?

- 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

y = x − 2
y = x − 2 + ln4
y = 2ln(x − 2)
y = 1
This is what it shows

- freckles

oh they want the linear approximation for f(2)

- freckles

they want the tangent line at x=2

- freckles

so just find the tangent line at x=2

- anonymous

When I derive 2lnx I get 2/x

- freckles

right now plug in 2

- freckles

\[y-f(2)=f'(2)(x-2) \\ y=f(2)+f'(2)(x-2)\]

- anonymous

I got 1 when I plugged it in

- freckles

ok so f'(2)=1
you still need to find f(2)

- freckles

\[y=f(2)+1(x-2) \\ y=f(2)+(x-2) \\ y=f(2)+x-2 \\ y=x-2+f(2)\]

- freckles

f(x)=2ln(x)
input 2 to find f(2)

- anonymous

1.39

- freckles

don't approximate
plug in 2

- freckles

then just use power rule for log

- freckles

\[a \ln(x)=\ln(x^a) \text{ power rule }\]

- freckles

\[f(2)=2\ln(2)=...\]

- anonymous

When I plugged that into my calculator I got 1.39

- 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

example
\[5\ln(2)=\ln(2^5)=\ln(32)\]

- freckles

you try
2ln(2)=what using power rule

- anonymous

So I would do ln(2^2)=ln(4)

- freckles

yes

- 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

x-2+ln4?

- freckles

yes

- freckles

y=x-2+ln(4)

- anonymous

Thank you so much!!!!

- freckles

np

Looking for something else?

Not the answer you are looking for? Search for more explanations.