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tangents
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derivative of this is lnx + 1
So the thing is that this tangent takes on many values, dependent on x. Since only give one particular x, I'd be tempted to use that. The problem is that ln(0) is undefined.
Let \( ( a, a \ln a) \) be the point where the tangent from (0,-3) touches the graph. \[ slope= \frac{a\ln a +3}{a-0}= \ln a +1\\ a \ln a +3 = a \ln a + a \\ a=3\\ \\ slope =\ln 3 + 1 \]
A quick sketch would save me some embarrassment
@telliott99 even that ln(0) is not defined, one can show that \[ \lim_{x\to 0} x \ln x =0 \]
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@eliassaab ,can u just explain how could u consider the point (a,alna)....?? means u took x,y both as a in y=xlnx.....how?
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