Asked by Tibor
Find the co-ordinates where f(x)intersects f'(x).
f(x)= ln(x+3)
f'(x)= -3+e^x
I rewrote f(x) so that it looks like: e^y=x+3 then I inserted f'(x): y=-3+e^x into f(x) and after a bit of working I got x=-10.6 which is not possible because ln of something negative is not possible.
f(x)= ln(x+3)
f'(x)= -3+e^x
I rewrote f(x) so that it looks like: e^y=x+3 then I inserted f'(x): y=-3+e^x into f(x) and after a bit of working I got x=-10.6 which is not possible because ln of something negative is not possible.
Answers
Answered by
Reiny
Wolfram shows 2 solutions, which verify when substituted.
http://www.wolframalpha.com/input/?i=+solve++ln%28x%2B3%29+%3D+-3%2Be%5Ex+
btw, I assume that your f ' (x) is not the derivative of the given f(x) .
http://www.wolframalpha.com/input/?i=+solve++ln%28x%2B3%29+%3D+-3%2Be%5Ex+
btw, I assume that your f ' (x) is not the derivative of the given f(x) .
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