Answer:
The absolute extrema of the function [tex]f(x) =x+e^{-3\cdot x}[/tex] on the domain [tex][-2,2][/tex] are:
Absolute minimum: [tex](x, f(x)) = (0.366, 0.700)[/tex]
Absolute maximum: [tex](x, f(x)) = (-2, 401.429)[/tex]
Step-by-step explanation:
Let [tex]f(x) =x+e^{-3\cdot x}[/tex] with interval in [tex][-2,2][/tex], we can check if the absolute extrema exist by applying First and Second Derivative Test, otherwise we must evaluate the function at lower and upper bounds.
First, we obtain the first derivative of the function, which is later equalized to zero and solved for [tex]x[/tex]:
[tex]1 -3\cdot e^{-3\cdot x} = 0[/tex]
[tex]3\cdot e^{-3\cdot x} = 1[/tex]
[tex]e^{-3\cdot x} = \frac{1}{3}[/tex]
[tex]-3\cdot x = \ln \frac{1}{3}[/tex]
[tex]x = -\frac{1}{3}\cdot \ln \frac{1}{3}[/tex]
[tex]x \approx 0.366[/tex]
Second, we evaluated the second derivative of given function at result above:
[tex]f''(x)= 9\cdot e^{-3\cdot x}[/tex]
[tex]f''(0.366) = 9\cdot e^{-3\cdot (0.366)}[/tex]
[tex]f''(0.366) \approx 3.002[/tex] (Absolute minimum)
The value of the absolute minimum is:
x = 0.366
[tex]f(0.366) =0.366+e^{-3\cdot (0.366)}[/tex]
[tex]f(0.366) \approx 0.700[/tex]
Third, we evaluate the function at each bound in the search for absolute maximum:
x = -2
[tex]f(-2) = -2+e^{-3\cdot (-2)}[/tex]
[tex]f(-2) \approx 401.429[/tex]
x = 2
[tex]f(2) = 2+e^{-3\cdot (2)}[/tex]
[tex]f(2) \approx 2.002[/tex]
The absolute extrema of the function [tex]f(x) =x+e^{-3\cdot x}[/tex] on the domain [tex][-2,2][/tex] are:
Absolute minimum: [tex](x, f(x)) = (0.366, 0.700)[/tex]
Absolute maximum: [tex](x, f(x)) = (-2, 401.429)[/tex]
