use the first and second derivatives to find the x-coordinates of all local maxima and minima of f(x)=8e^2x-2x. indicate which of your answers are local minima and which are local maxima.
To find the x-coordinates of local maxima and minima of the function f(x) = 8e^(2x) - 2x, we need to follow these steps:
Step 1: Find the first derivative of f(x) to obtain f'(x), and then find the second derivative to obtain f''(x).
Step 2: Set f'(x) = 0 and solve for x to find the critical points.
Step 3: Test each critical point by plugging it into f''(x).
- If f''(x) > 0, it is a local minimum.
- If f''(x) < 0, it is a local maximum.
Let's begin:
Step 1: Find the first and second derivatives of f(x):
f(x) = 8e^(2x) - 2x
Differentiating f(x) once:
f'(x) = (8e^(2x))'( - (2x)'
Differentiating e^(2x):
f'(x) = 8e^(2x) - 2
Differentiating f'(x) once more:
f''(x) = (8e^(2x) - 2)'
Differentiating e^(2x):
f''(x) = 16e^(2x)
Step 2: Find the critical points by setting f'(x) = 0:
8e^(2x) - 2 = 0
8e^(2x) = 2
e^(2x) = 2/8
e^(2x) = 1/4
Take the natural logarithm of both sides to eliminate the exponential:
2x = ln(1/4)
x = (1/2)ln(1/4)
x ≈ -0.6931
Step 3: Test each critical point using f''(x):
For x = -0.6931, let's plug it into f''(x):
f''(-0.6931) = 16e^(2(-0.6931))
f''(-0.6931) ≈ 16(0.5) ≈ 8
Since f''(-0.6931) > 0, this critical point is a local minimum.
Therefore, the x-coordinate of the local minimum is x ≈ -0.6931.
In this case, there is only one local minimum.