Find the linearization of f(x)=cos(x)-sin(2x) at x = 0.
L(x) = f(a) + f ' (a) (x-a) , where x = a
f(x) = cosx - sin 2x
f ' (x) = -sinx - 2cos 2x
L(x) = f(0) + f ' (0) (x-0)
= 1-0 + (0-2)(x-0)
= 1 - 2x
To find the linearization of a function at a specific point, we need to use the concept of the tangent line. The tangent line represents a linear approximation of the function near the given point.
To find the linearization of the function f(x) = cos(x) - sin(2x) at x = 0, we need to follow these steps:
Step 1: Determine the value of the function at the given point:
Evaluate f(0) by substituting x=0 into the function equation:
f(0) = cos(0) - sin(2*0)
f(0) = cos(0) - sin(0)
f(0) = 1 - 0
f(0) = 1
Step 2: Find the derivative of the function:
Compute the derivative of f(x) with respect to x. In this case, f(x) = cos(x) - sin(2x):
f'(x) = -sin(x) - 2cos(2x)
Step 3: Evaluate the derivative at the given point:
Compute f'(0) by substituting x=0 into the derivative equation:
f'(0) = -sin(0) - 2cos(2*0)
f'(0) = -sin(0) - 2cos(0)
f'(0) = -0 - 2(1)
f'(0) = -2
Step 4: Construct the equation of the tangent line:
The equation of a straight line can be represented in the form y = mx + b, where m is the slope of the line and b is the y-intercept.
Since the tangent line passes through the point (0, f(0)), we have the point (0, 1) from Step 1.
We also know the slope of the line is given by the derivative at the point, f'(0) = -2 from Step 3.
Using the point-slope form of a line, we can substitute these values into the equation:
y - y1 = m(x - x1), where (x1, y1) = (0, 1) and m = -2
Simplifying:
y - 1 = -2(x - 0)
y - 1 = -2x
y = -2x + 1
Step 5: The linearization of the function:
The linearization is given by the equation of the tangent line we obtained in Step 4:
L(x) = -2x + 1
Therefore, the linearization of f(x) = cos(x) - sin(2x) at x = 0 is L(x) = -2x + 1.