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?
[Math] Tangent of curve which passes through origin. Finding unknown variable
calculusderivatives
calculusderivatives
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?
Best Answer
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 $c = 1/a$.