Find and classify all local extreme values of each given function.
(a) g(x)=e^(2x)+e^(-x)
(b) g(x)=x^(2/3)*(x^(2)-4)
To find the local extreme values of a function, we need to find its critical points first. Critical points are the points where the derivative of the function is either zero or undefined.
Let's find the critical points for both functions:
(a) g(x) = e^(2x) + e^(-x)
1. Find the first derivative of g(x):
g'(x) = 2e^(2x) - e^(-x)
2. Set the derivative equal to zero and solve for x:
2e^(2x) - e^(-x) = 0
To solve this equation, let's rearrange it:
2e^(2x) = e^(-x)
Divide both sides by e^(-x):
2e^(3x) = 1
Now take the natural logarithm of both sides:
ln(2e^(3x)) = ln(1)
ln(2) + 3x = 0
Solve for x:
3x = -ln(2)
x = -ln(2)/3
So, x = -ln(2)/3 is the critical point.
(b) g(x) = x^(2/3)*(x^2 - 4)
1. Find the first derivative of g(x):
g'(x) = (2/3)x^(-1/3)*(x^2 - 4) + x^(2/3)*(2x)
2. Set the derivative equal to zero and solve for x:
(2/3)x^(-1/3)*(x^2 - 4) + x^(2/3)*(2x) = 0
To simplify this equation, we need to multiply both sides by 3 to eliminate the fraction:
2(x^2 - 4) + 3x(x)^(2/3) = 0
Expanding the terms:
2x^2 - 8 + 3x^(5/3) = 0
There is no simple algebraic way to solve this equation. We can use numerical methods, such as graphing or using a calculator, to find the approximate values of x.
Once you have the critical points, you can classify them as either maximum or minimum points by applying the second derivative test:
1. Find the second derivative of the function.
(a) The second derivative of g(x) = e^(2x) + e^(-x) is:
g''(x) = 4e^(2x) + e^(-x)
2. Substitute the critical point value of x into the second derivative.
(a) Let's substitute x = -ln(2)/3 into g''(x):
g''(-ln(2)/3) = 4e^(2(-ln(2)/3)) + e^(-(-ln(2)/3))
Simplify the expression to find the value.
(b) Use the same process for function g(x) = x^(2/3)*(x^2 - 4).
The second derivative test states that if the second derivative is positive at a critical point, the point corresponds to a local minimum. If the second derivative is negative, the point corresponds to a local maximum. If the second derivative is zero or undefined, the test is inconclusive.
By determining the sign of the second derivative at each critical point, you can classify whether it is a local minimum or maximum.