Find the local extrema of the function f(x) = (e^x^2+1)/(2x+1) defined for x > 0, and specify whether they are minima or maxima and give reasons why they are what you claim they are.
To find the local extrema of a function, we need to examine the critical points and determine their nature as minima or maxima using the First and Second Derivative Tests. Here's how you can do it step by step:
1. Find the derivative of the function f(x) with respect to x. Let's call this derivative f'(x). This step is necessary to identify the critical points where the extrema may occur.
Let's find the derivative f'(x):
f(x) = (e^(x^2)+1)/(2x+1)
To differentiate this function, we will use the Quotient Rule:
f'(x) = [((2x+1)(2x)(e^(x^2)+1)) - (e^(x^2)+1)(2)] / (2x+1)^2
Simplifying f'(x), we get:
f'(x) = (2x(e^(x^2)+1) - 2(e^(x^2)+1)) / (2x+1)^2
2. Find the critical points by setting f'(x) equal to zero and solving for x. Critical points are the x-values where the derivative is either zero or undefined.
Setting f'(x) equal to zero:
(2x(e^(x^2)+1) - 2(e^(x^2)+1)) / (2x+1)^2 = 0
Simplifying, we get:
2x(e^(x^2)+1) - 2(e^(x^2)+1) = 0
Dividing both sides by 2(e^(x^2)+1) and simplifying further:
x = 1
So the critical point is x = 1.
3. Determine whether the critical point is a local minimum or maximum. We will use the Second Derivative Test to do this.
To apply the Second Derivative Test, we need to find the second derivative f''(x).
Taking the derivative of f'(x):
f''(x) = [2(e^(x^2)+1) - 4x^2(e^(x^2)+1) - 4x(e^(x^2)+1)] / (2x + 1)^3
4. Substitute the critical point x = 1 into the second derivative f''(x).
f''(1) = [2(e^(1^2)+1) - 4(1)^2(e^(1^2)+1) - 4(1)(e^(1^2)+1)] / (2(1) + 1)^3
Simplifying further:
f''(1) = 2e + 2 / 3^3
f''(1) = (2e + 2) / 27
Since the second derivative f''(1) is positive, it implies that the function is concave up at the critical point x = 1. This indicates that the critical point is a local minimum.
Hence, the local extremum of the function f(x) = (e^x^2+1)/(2x+1) is a local minimum at x = 1.