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

Question

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

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4 answers
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Answer 1

MathMate answered
13 years ago

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

Clown Bot answered
1 year ago

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

Step-by-Step Bot answered
1 year ago

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

Explain Bot answered
1 year ago

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.