What is the linearization of f(x)=e^x at x=1?

Question

Generate an image that illustrates the concept of linearization. Depict a smoothly curved graph representing the exponential function (f(x) = e^x), with a clear marking at the point where x = 1. Around this point, show a tangent line that represents the linear approximation of the function at x = 1. Both the curve and the tangent line should be annotated with arrows, but without any text. Use vivid colors to distinguish between the curve, the point of tangency, and the tangent line.

What is the linearization of f(x)=e^x at x=1?

Answer 1

Linearization is the approximation of the derivative of the function by a tangent line at a particular point.

f(x)=e^x
f'(x)=e^x
at x=1, f'(1)=e
So the tangent of slope "e" passes through the point (1,e).
The equation of the tangent line is therefore
(y-e)=e(x-1)
y=e*x-e+e
=e*x

See figure at the link below, and note that the line y=e*x passes through the origin.

http://img607.imageshack.us/img607/4157/1291771397.png

Answer 2

Oh, so you want to unleash the mathematical clown in me! Alright, let's get the party started.

To find the linearization of f(x)=e^x at x=1, we need to use the concept of a tangent line. Imagine a unicycle rolling along the surface of the graph of f(x)=e^x. At the point (1, e^1), we place a brave circus performer on that unicycle, let's call him Uni-corny.

Uni-corny is so talented that he rides the tangent line to the graph of f(x)=e^x at x=1. Now, it turns out that the tangent line at that point is just the linearization of f(x)=e^x. How neat is that?

So, Uni-corny starts his ride on the tangent line, which we can express as y = f(1) + f'(1)(x - 1). Since f(x) = e^x, f'(x) = e^x. Plugging in x=1 yields f'(1) = e.

Now, to make it extra fun, let's put on our colorful math hat and write down the linearization: y = e^1 + e(x - 1), which simplifies to y = e + e(x - 1).

Ta-da! There you have it, the clown-approved linearization of f(x)=e^x at x=1. Now, enjoy the circus of mathematics! 🎪🤡

Answer 3

To find the linearization of a function f(x) at a specific point (x=a), we can use the formula:

L(x) = f(a) + f'(a)(x-a)

In this case, the function is f(x) = e^x, and we want to find the linearization at x=1.

Step 1: Find f(a)
Evaluate the function at x=1:
f(1) = e^1 = e

Step 2: Find f'(a)
Differentiate the function f(x) = e^x:
f'(x) = e^x

Evaluate the derivative at x=1:
f'(1) = e^1 = e

Step 3: Plug all values into the linearization formula
L(x) = f(a) + f'(a)(x-a)
L(x) = e + e(x-1)

Therefore, the linearization of the function f(x) = e^x at x=1 is L(x) = e + e(x-1).

Answer 4

To find the linearization of the function f(x) = e^x at x = 1, we need to calculate the equation of the tangent line to the graph of f(x) at x = 1.

The general equation of a linear function is y = mx + b, where m represents the slope of the line and b represents the y-intercept.

To find the slope, we can take the derivative of the function f(x) = e^x. The derivative of e^x with respect to x is simply e^x.

Now, let's find the slope of the tangent line at x = 1 by evaluating the derivative at that point: f'(1) = e^1 = e.

So, the slope of the tangent line at x = 1 is e.

To find the y-intercept of the tangent line, we substitute the point (1, f(1)) into the equation of the tangent line.

Since f(1) = e^1 = e, the y-coordinate of the point is e.

Now we have the slope (e) and the y-intercept (e) of the tangent line.

Plugging these values into the equation of a line, we get: y = ex + e.

Therefore, the linearization of f(x) = e^x at x = 1 is y = ex + e.