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If the tangent to the graph of $y = e^{ax}$, $a \ne 0$, at $x = c$ passes through the origin, then $c$ is equal to?

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  • $\begingroup$ Can you share what you've tried, and explain what's giving you trouble? For example: Do you know how the tangent line is related to the derivative? $\endgroup$ – user61527 Jan 2 '14 at 4:00
  • $\begingroup$ Sure. So I thought maybe you had to derive y=e^ax so i got gradient of tangent = a.e^ax. (but x=c), so it would be a.e^a(c) the equation of tangent is at (0,0) y=mx+b 0=a.e^a(c)(0) + b and from here i have no idea what i'm doing, because i keep getting zero and im not sure if thats even right. I honestly just have no clue how to approach this. the options to choose from are 1,0,a, 1/a, -1/a $\endgroup$ – confused Jan 2 '14 at 4:03
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The general equation for the tangent line to $f$ at the point $c$ is

$$y = f'(c) (x - c) + f(c)$$

Since the derivative of $e^{ax}$ is $a e^{ax}$, this takes the form

$$y = ae^{ac} (x - c) + e^{ac}$$

This needs to pass through the origin, so we need

$$0 = -ca e^{ac} + e^{ac} \implies e^{ac} (1 - ac) = 0$$

Since $e^{ac} \ne 0$ for any choice of $c$, we see that $c = 1/a$.

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  • $\begingroup$ wow, that was so simple. :O i feel like i've just discovered the most amazing general equation. HAHA. Thank you so much! So do i use this formula for whenever x= ??(so any point on the curve) ?? $\endgroup$ – confused Jan 2 '14 at 4:08
  • $\begingroup$ @confused Yes: This equation gives the value of the tangent line at any point $x$; by altering $c$, you can change where the tangent line intersects $f$. $\endgroup$ – user61527 Jan 2 '14 at 4:09
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Let's start by taking the derivative of $e^{ax}$ at $x=c$. Doing this, we get $ae^{ac}$. This is going to be the slope of the tangent line at $x=c$. The $y$-value of the tangent line at $x=c$ is then just the $y$-value of our original curve at $x=c$---i.e. $e^{ac}$. Thus, the tangent line (in point-slope form) is $$ y=ae^{ac}(x-c) + e^{ac}.$$

Now, we're told that this line passes through the origin. That means when we plug in $x=0, y=0$ to the above equation for the tangent line, we should get equality. So, $$ 0 = -ace^{ac} + e^{ac}.$$ Thus, cancelling out $e^{ac}$ and solving for $c$, we get $c=1/a$.

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  • $\begingroup$ I think you mean $c = 1/a$. $\endgroup$ – user61527 Jan 2 '14 at 5:09

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