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Show that the $x$ -intercept of the tangent line to the curve $y=e^{a x}$ at the point at which $x=c$ is $\frac{a c-1}{a} .$ In particular, show the $x$ -intercept of the tangent line to the curve $y=e^{x}$ is always one unit to the left of the point of tangency.

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 4

The Derivative of the Exponential Function

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Lectures

01:05

Show that the $x$ -interce…

02:11

Show that the $x$-intercep…

00:40

(a) Find the equation of a…

04:31

show that the sum of the x…

02:37

Show that the sum of the $…

04:09

05:38

Show that any two tangent …

Okay here we want to start with the curve, Y equals E. To the X. So here let's give a rough sketch of our curve here's thanks. Here's why. And we don't know if the expert is going to be increasing or decreasing. Right? If we have Y equals E. To the X. If a is positive then it's gonna be an increasing exponentially. And if A is negative it's going to be decreasing but it's kind of symmetric. So let's just try to get an idea of what's going to happen by drawing an increasing exponential. And then we're going to have some point X equal sees. Let's just give an example here, that's our point. And then we're taking the tangent line to this point and we're saying that it intersects the X axis and supposedly intersect according to this question at the value X equals AC -1, divided by a. Okay, so to verify this, we first need to find the equation of the tangent line at a general point X equals C. So to do that we need the derivative of our curve. So why prime here get me each of the A. X. To take through of an exponential. It's just the exponential. And then by the chain rule we have to multiply by the coefficient of X. Or the derivative of a X. Which is just A. Then at X equals C. Well I mean sees there's a general point. Uh So we're gonna get, just all have to do is plug it in a C. And then we have to go ahead and got the equation of the tangent line. This is just the slope of the tangent line Y. Prime. And to do that we need the arbitrary point here. So the coordinates of the point of tangent C. It has X coordinate C. And so that means it's why coordinated by this relationship is going to be easy to A C. It's the Y value get when you substitute in X equals C. And so our equation of our line is going to be why minus that Y coordinate E. To the A. C. Physical to the slope slope of this we just computed and then times X minus the X coordinate. Okay, so that's that's the equation of the tension line at an arbitrary point on this graph. And we want to find the Y or sorry, the x intercept. Well the x intercept is the value of X when y is zero. So in other words we're gonna substitute in Y equals zero. So that just leaves this left and everything else stays the same and we want to solve for X here. Okay, well one thing we can do as we notice that each of the A. C. Appears as a factor on both sides of the equation so we can cancel those out. And what we're left with is negative. one equals a X minus C. Now there's a couple of ways to solve this. We can multiply it out first. So multiply or distribute the aid to both terms and we get this, then add a C. On both sides to isolate X, so a c minus one equals X, and then divide by A. On both sides and there you go. That is the x intercept of the tangent line, and you can see it. Indeed, uh is what the question is asking for.

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