f(x)=x+1/x Find all critical numbers. Find where the function is increasing and decreasing. Find critical points and identify each as a relative maximum, relative minimum, or neither.
Find second order critical numbers and tell where the graph is concave up and where it is concave down.
I'm on the last question now, but i don't know how to find the zeros of the second derivative because i got 2/x^3..
That just means that f"(x) is never zero. There are no points of inflection.
visit wolframalpha.com and type in your function. It will show you the graph, and I think your questions will be answered.
To find the critical numbers of a function, you need to find the values of x at which the derivative of the function is equal to zero or is undefined. Let's go step by step to find the critical numbers, determine where the function is increasing or decreasing, and identify the relative maximum, relative minimum, and inflection points.
1. Find the derivative of f(x):
f'(x) = 1 - 1/x^2
2. Set the derivative equal to zero and solve for x:
1 - 1/x^2 = 0
To solve this equation, subtract 1 from both sides:
-1/x^2 = -1
Take the reciprocal of both sides:
x^2 = 1
Take the square root of both sides:
x = ±1
So the critical numbers are x = 1 and x = -1.
3. Determine where the function is increasing or decreasing:
To determine where the function is increasing or decreasing, we can use the first derivative test. Analyze the sign of the derivative in the intervals of the critical numbers.
For x < -1:
Choose x = -2 to test:
f'(-2) = 1 - 1/(-2)^2 = 1 - 1/4 = 3/4 > 0
Since the derivative is positive, the function is increasing in this interval.
For -1 < x < 1:
Choose x = 0 to test:
f'(0) = 1 - 1/0^2 = 1 - Undefined
The derivative is undefined at x = 0, so we need to analyze further.
For x > 1:
Choose x = 2 to test:
f'(2) = 1 - 1/2^2 = 1 - 1/4 = 3/4 > 0
The derivative is positive, so the function is increasing in this interval.
4. Identify the relative maximum, relative minimum, and inflection points:
To identify the relative maximum, relative minimum, and inflection points, we need to analyze the second derivative and determine where it changes sign.
The second derivative of f(x) is the derivative of the first derivative:
f''(x) = d/dx (f'(x)) = d/dx (1 - 1/x^2) = 2/x^3
Now, we need to find the zeros of the second derivative to determine the points of inflection.
To find the zeros of 2/x^3, set the second derivative equal to zero and solve for x:
2/x^3 = 0
Since 2 is not zero, there are no solutions for x.
Since the second derivative does not change sign or have any zeros, there are no points of inflection.
Lastly, we need to determine where the graph is concave up or concave down. You can use the second derivative test to analyze the sign of the second derivative.
For x < 0:
Choose x = -1 to test:
f''(-1) = 2/(-1)^3 = -2 < 0
The second derivative is negative, so the graph is concave down in this interval.
For x > 0:
Choose x = 1 to test:
f''(1) = 2/(1)^3 = 2 > 0
The second derivative is positive, so the graph is concave up in this interval.
To summarize:
- The critical numbers are x = 1 and x = -1.
- The function is increasing for x < -1 and x > 1.
- The function is decreasing for -1 < x < 1.
- There are no relative maximum, relative minimum, or inflection points.
- The graph is concave down for x < 0 and concave up for x > 0.
I hope this explanation helps you understand the process of finding critical numbers, determining where a function is increasing/decreasing, identifying relative maximum/minimum, and analyzing concavity.