Find the linear approximation of the function.
g(x) = ^3sqrt(1 + x) at a = 0.
g(x)≈
Use to approximate these numbers (round to 3rd decimal)
3^sqrt(.95)
3^sqrt(1.1)
g'(x) = 1/3 (1+x)^(-2/3)
g'(0) = 1/3
g(0) = 1
so the linear approximation is g(x) ≈ 1 + 1/3 x
Why did the function go to school? Because it wanted to get a degree in approximation!
Now, let's find the linear approximation of the function g(x) = ^3sqrt(1 + x) at a = 0.
To do this, we'll use the formula for a linear approximation:
L(x) = f(a) + f'(a)(x - a)
First, let's find f(a):
f(a) = ^3sqrt(1 + 0) = ^3sqrt(1) = 1.
Now, let's find f'(x):
f'(x) = (1/3)(1 + x)^(-2/3).
Next, we need to find f'(a):
f'(0) = (1/3)(1 + 0)^(-2/3) = (1/3)(1)^(-2/3) = 1/3.
Finally, let's plug everything into the formula:
L(x) = 1 + (1/3)(x - 0) = 1 + x/3.
Now, let's use this linear approximation to approximate the numbers you provided:
Approximation of 3^sqrt(0.95):
L(0.95) = 1 + 0.95/3 = 1 + 0.316 = 1.316.
Approximation of 3^sqrt(1.1):
L(1.1) = 1 + 1.1/3 = 1 + 0.367 = 1.367.
So, the linear approximations of 3^sqrt(0.95) and 3^sqrt(1.1) are approximately 1.316 and 1.367, respectively.
To find the linear approximation of the function g(x) = ∛(1 + x) at a = 0, we can use the tangent line equation, which can be written as:
L(x) = f(a) + f'(a)(x - a)
Where f(a) is the value of the function at a, f'(a) is the derivative of the function at a, and (x - a) is the difference between x and a.
First, let's find f(a) and f'(a):
f(a) = ∛(1 + 0) = ∛1 = 1
To find f'(a), we need to take the derivative of g(x) with respect to x:
g(x) = ∛(1 + x)
g'(x) = (1/3)(1 + x)^(-2/3)
Now, substitute a = 0 into g'(x):
g'(a) = (1/3)(1 + 0)^(-2/3) = (1/3)(1)^(-2/3) = (1/3)(1) = 1/3
Now, we can find the linear approximation of g(x):
L(x) = f(a) + f'(a)(x - a)
L(x) = 1 + (1/3)(x - 0)
L(x) = 1 + (1/3)x
L(x) = (1/3)x + 1
Using this linear approximation, we can approximate the numbers:
3^∛(0.95) ≈ 3^L(0.95) ≈ 3^((1/3)(0.95) + 1) ≈ 3^(0.316667 + 1) ≈ 3^1.316667 ≈ 2.814
3^∛(1.1) ≈ 3^L(1.1) ≈ 3^((1/3)(1.1) + 1) ≈ 3^(0.366667 + 1) ≈ 3^1.366667 ≈ 3.076
Therefore, the approximations are:
3^∛(0.95) ≈ 2.814
3^∛(1.1) ≈ 3.076
To find the linear approximation of a function at a given point, we can use the tangent line to the function at that point. The equation of a tangent line is given by the point-slope form, y - y₁ = m(x - x₁), where (x₁, y₁) is the point of tangency and m is the slope of the tangent line.
To find the slope of the tangent line (m), we can use the derivative of the function evaluated at the given point (a). In this case, we need to find the derivative of g(x) = ^3√(1 + x) and evaluate it at a = 0.
Let's find the derivative of g(x) first:
g(x) = ^3√(1 + x)
Taking the derivative of g(x) using the chain rule:
g'(x) = (1/3) * (1 + x)^(-2/3) * 1
Simplifying this expression:
g'(x) = (1/3√(1 + x)^2)
Now, plug in a = 0 to find the slope at that point:
g'(0) = (1/3√(1 + 0)^2)
g'(0) = (1/3√(1^2))
g'(0) = 1/3
The slope of the tangent line is 1/3. Now we need to find the y-coordinate of the point of tangency, which is g(0).
g(0) = ^3√(1 + 0)
g(0) = ^3√1
g(0) = 1
So the point of tangency is (0, 1). Now, we can substitute these values into the point-slope form to find the equation of the tangent line.
(y - 1) = (1/3)(x - 0)
y - 1 = 1/3x
y = 1/3x + 1
This is the equation of the tangent line to the function g(x) = ^3√(1 + x) at a = 0.
Using this linear approximation, g(x)≈ 1/3x + 1, we can approximate given numbers by plugging them into the equation.
To approximate 3^√(0.95):
Plug x = 0.95 into the linear approximation equation:
g(0.95) ≈ 1/3(0.95) + 1
g(0.95) ≈ 0.31667 + 1
g(0.95) ≈ 1.3167 (rounded to 3 decimal places)
To approximate 3^√(1.1):
Plug x = 1.1 into the linear approximation equation:
g(1.1) ≈ 1/3(1.1) + 1
g(1.1) ≈ 0.36667 + 1
g(1.1) ≈ 1.3667 (rounded to 3 decimal places)
Therefore, the linear approximations of 3^√(0.95) and 3^√(1.1) are approximately 1.317 and 1.367, respectively (rounded to 3 decimal places).