Let f be the function given by f(x)= x^2+4x-8. The tangent line to the graph at x=2 is used to approximate values of f. For what value(s) of x is the tangent line approximation twice that of f?
2X+4
2(2)+4
4+4
F(2)=8
To find the value(s) of x where the tangent line approximation is twice that of f(x), we need to follow these steps:
Step 1: Find the derivative of the function f(x).
The derivative of f(x) = x^2 + 4x - 8 can be found by applying the power rule:
f'(x) = 2x + 4
Step 2: Find the slope of the tangent line at x = 2.
Evaluate the derivative function at x = 2:
f'(2) = 2(2) + 4
f'(2) = 8
So, the slope of the tangent line to the graph of f at x = 2 is 8.
Step 3: Write the equation of the tangent line at x = 2.
We'll use the point-slope form of a linear equation, where the point (x1, y1) lies on the line and m is the slope of the line.
The point (2, f(2)) lies on the tangent line, so we substitute these values into the point-slope form:
y - f(2) = 8(x - 2)
Step 4: Set up an equation to find the x-values where the tangent line is twice that of f.
We need to find the x-values where the tangent line approximation is twice that of f(x). Let's call these values x_2.
To find x_2, we set up the equation:
2f(x_2) = y
Step 5: Find the values of x that satisfy the equation.
Substitute f(x_2) from step 4 into the equation of the tangent line to get:
2(f(x_2)) = 8(x_2 - 2) + f(2)
Simplify and rearrange:
2(x_2^2 + 4x_2 - 8) - 8(x_2 - 2) = x_2^2 + 4x_2 - 8
Expand and collect like terms:
2x_2^2 + 8x_2 - 16 - 8x_2 + 16 = x_2^2 + 4x_2 - 8
Simplify and combine like terms:
x_2^2 = 0
Since x_2^2 = 0, the only solution is x_2 = 0.
Therefore, the tangent line approximation is twice that of f(x) at x = 0.