consider the following function:
f(x)=sqrt(50-x)
find the linear approximation of f(x) at a=1
To find the linear approximation of a function at a given point, we use the concept of the tangent line. The tangent line at a point on a graph approximates the behavior of the function near that point.
In this case, we want to find the linear approximation of the function f(x) = √(50 - x) at a = 1.
To find the linear approximation, we need the value of the function at the point a = 1, as well as the slope of the tangent line at that point. We can compute both of these values:
1. Value of the function at a = 1:
Plugging in a = 1 into the function, we get f(1) = √(50 - 1) = √49 = 7. So, the value of f(x) at a = 1 is 7.
2. Slope of the tangent line at a = 1:
To find the slope, we need to find the derivative of the function. In this case, f(x) = √(50 - x), so we can use the power rule for differentiation.
Taking the derivative of f(x), we get:
f'(x) = -1/(2√(50 - x))
Now, plugging in a = 1 into the derivative, we get:
f'(1) = -1/(2√(50 - 1)) = -1/(2√49) = -1/14
Therefore, the slope of the tangent line at a = 1 is -1/14.
Now that we have the value of the function at a = 1 (7) and the slope of the tangent line at a = 1 (-1/14), we can write the equation of the tangent line:
y = f(a) + f'(a)(x - a)
Plugging in the values we found, the equation becomes:
y = 7 - (1/14)(x - 1)
This is the linear approximation of the function f(x) = √(50 - x) at a = 1.