Find the local linearization of f(x) near x=0 and use it to approximate the value of a. f(x)=sqrt((b^2)+x) a=sqrt(26)
To find the local linearization of f(x) near x=0, we first need to compute the first derivative of f(x).
Given: f(x) = √(b^2 + x)
To find the first derivative of f(x), we can use the chain rule.
Step 1: Find the derivative of the expression inside the square root.
Let g(x) = b^2 + x
g'(x) = 1
Step 2: Apply the chain rule to find the derivative of f(x).
f'(x) = 1 / (2 * √(b^2 + x)) * 1
The local linearization of f(x) near x=0 is given by the equation:
L(x) = f(0) + f'(0)(x - 0)
Now, let's substitute the values into the equation:
L(x) = f(0) + f'(0)(x)
To approximate the value of a = √26 using the local linearization, we need to find the value of L(x) when x = a - 0.
L(a - 0) = f(0) + f'(0)(a - 0)
Now, substitute the values into the equation:
L(a - 0) = f(0) + f'(0)(a)
L(a - 0) = √(b^2 + 0) + (1 / (2 * √(b^2 + 0))) * (a)
L(a - 0) = √b^2 + a / (2 * √b^2)
Finally, substitute a = √26 into the equation:
L(√26 - 0) = √b^2 + √26 / (2 * √b^2)
Now, you can simplify the expression to find the approximation of a using the local linearization.