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Mathematics 51 Online
OpenStudy (anonymous):

find the equation of the tangent line to the graph of y=f(x) at x=xknot . #) f(x)=ln x; x=e^-1 answer is "y=ex-2", but how do you do that? I need to know the steps.

OpenStudy (anonymous):

yo ninjanick, were all here to help, but your question honestly isn't worded correctly at all. we don't know what you are asking.

OpenStudy (anonymous):

find the tangent line to the graph

OpenStudy (anonymous):

you didn't give us a graph or equation. is the equation ln(x) or e^-1. and at x=xknot doesn't mean anything either.

OpenStudy (anonymous):

here is the function f(x)=ln(x) and x=e^-1, answer is suppose to be "y=ex-2", so they want us to find the equation of the tangent line of that function.

OpenStudy (anonymous):

if the function is f(x) = ln(x) and x = e^-1, then it is not a function, it is simply ln(1/e), which is a number, not a function. Do you want the equation of the tangent line of f(x) = ln(x) at the point x=1/e?

OpenStudy (dumbcow):

\[f(x) = \ln x\] \[f'(x) = \frac{1}{x}\] \[x = e^{-1} \rightarrow f(e^{-1})=\ln(e^{-1}) = -1\] \[m = f'(e^{-1}) = \frac{1}{e^{-1}} = e\] \[y = mx + b\] \[\rightarrow -1 = e*(1/e) + b\] \[b = -2\] \[\rightarrow y = e*x - 2\]

OpenStudy (anonymous):

If that is the case, take derivative of ln(x) = i / x. At 1/e, we get the derivative = e.

OpenStudy (anonymous):

yep there u go hope thats what ur asking for

OpenStudy (anonymous):

thanks dumbcow, but you should be named smartcow.

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