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anonymous
 one year ago
???
anonymous
 one year ago
???

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freckles
 one year ago
Best ResponseYou've already chosen the best response.3you 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: yf(e)=f'(e)(xe) y=f(e)+f'(e)(xe) so f(x) is approximately f(e)+f'(e)(xe)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I would plug in values into that?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you can use that approximation equation to approximate f(2) f(2) is approximately f(e)+f'(e)(2e)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you still need to find f(e) and f'(e) though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would I find those. And would it help if i showed the answer choices?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3there are choices for approximating f(2)?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3f(e) can bound by pluggin in e into f(x)=2ln(x) f'(x) needs to be found before finding f'(e)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y = x − 2 y = x − 2 + ln4 y = 2ln(x − 2) y = 1 This is what it shows

freckles
 one year ago
Best ResponseYou've already chosen the best response.3oh they want the linear approximation for f(2)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3they want the tangent line at x=2

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so just find the tangent line at x=2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When I derive 2lnx I get 2/x

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[yf(2)=f'(2)(x2) \\ y=f(2)+f'(2)(x2)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 1 when I plugged it in

freckles
 one year ago
Best ResponseYou've already chosen the best response.3ok so f'(2)=1 you still need to find f(2)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[y=f(2)+1(x2) \\ y=f(2)+(x2) \\ y=f(2)+x2 \\ y=x2+f(2)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3f(x)=2ln(x) input 2 to find f(2)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3don't approximate plug in 2

freckles
 one year ago
Best ResponseYou've already chosen the best response.3then just use power rule for log

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[a \ln(x)=\ln(x^a) \text{ power rule }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When I plugged that into my calculator I got 1.39

freckles
 one year ago
Best ResponseYou've already chosen the best response.3yes 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
 one year ago
Best ResponseYou've already chosen the best response.3example \[5\ln(2)=\ln(2^5)=\ln(32)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you try 2ln(2)=what using power rule

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I would do ln(2^2)=ln(4)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3the tangent line at x=a looks like this yf(a)=f'(a)(xa) so we has 2 instead of a yf(2)=f'(2)(x2) you found f(2) as ln(4) and f'(2) as 1 yln(4)=1(x2) yln(4)=x2 add ln(4) on both sides and you should see clearly see your answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much!!!!
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