Question

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.

Answer 1

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.

Answer 2

find the tangent line to the graph

Answer 3

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.

Answer 4

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.

Answer 5

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?

Answer 6

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

Answer 7

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

Answer 8

yep there u go hope thats what ur asking for

Answer 9

thanks dumbcow, but you should be named smartcow.