Find the linear approximation of the function g(x)=cube root(1+x) at a=0.
Assuming you meant x=0,
g'(x) = 1/3 (1+x)^(-2/3)
g'(0) = 1/3
So, the line going through (0,1) with slope 1/3 is ?
See
http://www.wolframalpha.com/input/?i=plot+y%3D%281%2Bx%29^%281%2F3%29%2C+y%3Dx%2F3%2B1+for+-1%3C%3Dx%3C%3D2
Well, finding a linear approximation is like finding a good celebrity impersonator. You want someone who can capture the essence of the real person, but you know they're not quite the same. Similarly, a linear approximation captures the essence of a function near a certain point, but it's not an exact match.
To find the linear approximation of g(x) = ³√(1+x) at a = 0, we'll use a tangent line to approximate the function.
First, let's find the derivative of g(x). Since g(x) = ³√(1+x), we can use the chain rule to find that g'(x) = (1/3)(1+x)^(-2/3).
Now, let's evaluate g'(0). Plugging in x = 0 into the derivative, we get g'(0) = (1/3)(1+0)^(-2/3) = 1/3.
The equation of a tangent line is y = f(a) + f'(a)(x - a). Plugging in a = 0, (0) = ³√1 = 1, and '(0) = 1/3, we get y = 1 + (1/3)x.
Therefore, the linear approximation of g(x) = ³√(1+x) at a = 0 is y = 1 + (1/3)x. And just like an impersonator, it does a pretty good job imitating the function near a = 0, but you wouldn't mistake it for the real thing!
To find the linear approximation of a function at a specific point, we use the equation:
L(x) = f(a) + f'(a)(x - a)
where L(x) is the linear approximation, f(a) is the value of the function at the point a, f'(a) is the derivative of the function at the point a, and (x - a) is the difference between the given point and the point of approximation.
In this case, the function is g(x) = ∛(1 + x) and the point of approximation is a = 0.
Step 1: Find f(a)
To find f(a), substitute the value of a = 0 into the function g(x):
g(a) = g(0) = ∛(1 + 0) = ∛1 = 1
So, f(a) = 1.
Step 2: Find f'(a)
To find f'(a), differentiate the function g(x) with respect to x:
g'(x) = d/dx(∛(1 + x))
= (1/3)(1 + x)^(-2/3) * 1
= (1/3)(1 + x)^(-2/3)
Now, substitute the value of a = 0 into g'(x):
g'(a) = g'(0) = (1/3)(1 + 0)^(-2/3) = (1/3)
So, f'(a) = 1/3.
Step 3: Substitute the values into the linear approximation formula
Using the linear approximation formula:
L(x) = f(a) + f'(a)(x - a)
Substitute f(a) = 1, f'(a) = 1/3, and a = 0:
L(x) = 1 + (1/3)(x - 0)
= 1 + (1/3)x
Therefore, the linear approximation of the function g(x) = ∛(1 + x) at a = 0 is L(x) = 1 + (1/3)x.
To find the linear approximation of a function at a specific point, we can use the formula for linear approximation:
L(x) = f(a) + f'(a)(x-a)
Where:
- L(x) is the linear approximation function
- f(a) is the value of the function at the point a
- f'(a) is the derivative of the function evaluated at a
- x is the variable of the function
In this case, we want to find the linear approximation of the function g(x) = cube root(1+x) at the point a = 0.
Step 1: Calculate the value of f(a)
In the given function, (a) = cube root(1+0) = cube root(1) = 1. Therefore, the value of f(a) is 1.
Step 2: Calculate the derivative of the function f'(a)
To find the derivative of g(x) = cube root(1+x), we can use the chain rule. Let's differentiate step by step:
- Let u = 1 + x
- Let y = u^(1/3)
Now, taking the derivative of y with respect to u,
dy/du = (1/3)u^(-2/3)
Next, taking the derivative of u with respect to x,
du/dx = 1
Lastly, using the chain rule,
dy/dx = (1/3)u^(-2/3) * du/dx = (1/3)(1 + x)^(-2/3)
Therefore, '(a) = (1/3)(1 + 0)^(-2/3) = (1/3)(1)^(-2/3) = 1/3.
Step 3: Substitute the values into the formula
Using the formula for linear approximation, L(x) = f(a) + f'(a)(x-a), we can substitute the values we found:
L(x) = f(a) + f'(a)(x-a)
L(x) = 1 + (1/3)(x-0)
L(x) = 1 + (1/3)x
Therefore, the linear approximation of the function g(x) = cube root(1+x) at a=0 is L(x) = 1 + (1/3)x.