f(x)=4-x^2-ln((1/2)x+1) find the tangent line approximation of f(-0.2)
To find the tangent line approximation of a function at a specific point, we need to find the slope of the tangent line at that point and also the y-coordinate of the point where the tangent line intersects the y-axis.
Let's start by finding the slope of the tangent line at x = -0.2. The slope of a tangent line can be found using the derivative of the function at that point.
Step 1: Find the first derivative of f(x)
Let's find the derivative of f(x) = 4 - x^2 - ln((1/2)x + 1).
First, find the derivative of each term:
f'(x) = d/dx (4) - d/dx (x^2) - d/dx (ln((1/2)x + 1))
The derivative of a constant is zero, so the first term becomes 0.
The derivative of x^2 is 2x.
The derivative of ln(u) is 1/u * du/dx, where u is the argument of the natural logarithm. So the third term becomes (1/((1/2)x + 1)) * d/dx ((1/2)x + 1).
However, since the derivative of 1/2 is zero, the third term becomes 0.
Therefore, f'(x) = 0 - 2x - 0 = -2x.
Step 2: Evaluate the derivative at x = -0.2
Substitute x = -0.2 into f'(x):
f'(-0.2) = -2(-0.2) = 0.4.
Now we have the slope of the tangent line, which is 0.4.
Step 3: Find the y-coordinate where the tangent line intersects the y-axis
To find the y-coordinate where the tangent line intersects the y-axis, we substitute x = -0.2 into the original function f(x):
f(-0.2) = 4 - (-0.2)^2 - ln((1/2)(-0.2) + 1).
Evaluating this expression will give us the y-coordinate at x = -0.2.
Step 4: Substitute the values into the equation of a line
Now that we have the slope (0.4) and the y-coordinate (-0.28249) where the tangent line intersects the y-axis, we can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1),
where (x1, y1) is the point of tangency.
Using (x1, y1) = (-0.2, f(-0.2)) = (-0.2, -0.28249), and m = 0.4, we have:
y + 0.28249 = 0.4(x + 0.2).
Simplifying this equation will give you the tangent line approximation of f(-0.2).