f(x)=(4+x)^1/2= 2(1+x/4)^1/2=2(1+x/8)
check that.
check that.
The linearization approximation formula is given by:
f(x) ≈ f(a) + f'(a)(x - a)
In this case, let's take a = 0. So, a = 0 and f(a) = f(0) = (1/√(4+0)) = 1/2.
Next, we need to find f'(x), which is the derivative of f(x). Let's differentiate f(x) with respect to x:
f(x) = (1/√(4+x))
Using the chain rule,
f'(x) = (-1/2)(4+x)^(-3/2)
Now, we can evaluate f'(a) at x = 0:
f'(0) = (-1/2)(4+0)^(-3/2) = -1/8
Finally, we can use the formula for linear approximation to find the approximation for f(x) near x = 0:
f(x) ≈ f(a) + f'(a)(x - a)
Substituting the values we found:
f(x) ≈ 1/2 + (-1/8)(x - 0)
Simplifying this expression, we get:
f(x) ≈ 1/2 - (1/8)x
Therefore, the linear approximation for the function f(x) = (1/√(4+x)) for values of x near zero is approximately given by f(x) ≈ 1/2 - (1/8)x.
1. Determine the value of k:
In our case, k is the exponent of the binomial (1+x)^k. In the function f(x), the term (4+x) appears in the denominator, so we can rewrite it as (1+x)^(-1/2). Thus, k is equal to -1/2.
2. Compute the derivative of f(x):
We need to evaluate the derivative of f(x) with respect to x. In this case, f(x) = (1/√(4+x)), so we need to apply the chain rule to differentiate it. The derivative is given by:
f'(x) = -(1/2)(4+x)^(-3/2)
3. Evaluate the function and its derivative at x = 0:
We substitute x = 0 into f(x) and f'(x) to get the values at x = 0:
f(0) = (1/√4) = 1/2
f'(0) = -(1/2)(4)^(-3/2) = -1/4
4. Use the linearization approximation formula:
The formula is given as: f(x) ≈ f(a) + f'(a)(x - a)
In our case, a = 0 since we are approximating for values of x near zero. Substituting the known values, the linearization approximation is:
f(x) ≈ f(0) + f'(0)(x - 0)
f(x) ≈ 1/2 + (-1/4)(x)
Simplifying, we get: f(x) ≈ 1/2 - x/4
Therefore, for values of x near zero, an approximation for the function f(x) = (1/√(4+x)) can be given by f(x) ≈ 1/2 - x/4.