Find the linear approximation of the function f(x) = √16-x at a = 0
What is a? It does not appear in your formula for f(x)
Do you mean
f(x) = (sqrt16) - x or sqrt(16-x)?
The first case is trivial: 4 -x.
For the second case, use the Taylor series formula.
At x=0, the derivative of f(x) is
f'(x) =
(-1/2)/(16-x)^1/2 = -1/8
In linear approximation,
f(x) = f(0) + f'(0)*x
= 4 - (x/8)
Well, since you asked for the linear approximation of √16-x at a = 0, let's give it a try. To find the linear approximation, we start with the equation of a line: y = mx + b. In this case, y represents the function value, x represents the input value, m represents the slope, and b represents the y-intercept.
First, we need to find the slope (m) of the tangent line to the function at x = 0. To do this, we take the derivative of f(x) = √16-x:
f'(x) = -1/(2√16-x).
We then plug in x = 0 to get the slope:
m = -1/(2√16-0) = -1/(2√16) = -1/(2 * 4) = -1/8.
Now, we need to find the y-intercept (b) by evaluating the function at x = 0:
f(0) = √16-0 = √16 = 4.
So, our linear approximation of the function f(x) = √16-x at a = 0 is:
y = -1/8 * x + 4.
I hope this approximation is good enough for you, or else I'll have to bring in my clown wig and juggle some equations!
To find the linear approximation of the function f(x) = √16-x at a = 0, we need to use the formula for linear approximation:
L(x) = f(a) + f'(a)(x - a)
First, let's find f(a) and f'(a). Since a = 0, we have:
f(a) = f(0) = √16-0 = √16 = 4
Next, let's find f'(a), which is the derivative of f(x):
f'(x) = d/dx (√16-x)
= (-1/2)*(16-x)^(-1/2)*(-1)
= (1/2)*(16-x)^(-1/2)
Plugging in a = 0:
f'(a) = f'(0) = (1/2)*(16-0)^(-1/2) = (1/2)*16^(-1/2) = 1/8
Now we can substitute these values back into the linear approximation formula:
L(x) = f(a) + f'(a)(x - a)
= 4 + (1/8)(x - 0)
= 4 + (1/8)x
= (1/8)x + 4
Therefore, the linear approximation of the function f(x) = √16-x at a = 0 is L(x) = (1/8)x + 4.
To find the linear approximation of a function at a given point, we can use the concept of the tangent line.
First, let's find the derivative of the function f(x) = √(16 - x). Differentiating with respect to x, we have:
f'(x) = -1/(2√(16 - x))
Next, we can evaluate the derivative at the point a = 0:
f'(0) = -1/(2√16) = -1/8
The slope of the tangent line at x = 0 is -1/8.
Now, let's find the function value at x = a = 0:
f(a) = f(0) = √(16 - 0) = √16 = 4
So, the function value at x = 0 is 4.
The equation for the tangent line to the function f(x) at x = 0 can be written in the point-slope form:
y - f(a) = f'(a)(x - a)
Substituting the values we found, we get:
y - 4 = -1/8(x - 0)
Simplifying:
y - 4 = -1/8x
Re-arranging the equation to make y the subject:
y = -1/8x + 4
This is the equation of the linear approximation of the function f(x) = √(16 - x) at x = 0.