Find the largest open intervals where the function is concave upward.
f(x)=x^4-8x^2
f" = 12x^2-16
f">0 where |x|>2/√3
That is, (-∞,-2/√3) (2/√3,∞)
Both those are intervals extending to infinity. Question is ill-posed.
Well, let's analyze the function f(x) = x^4 - 8x^2 and see where it's concave upward. To determine the concavity, we need to find the second derivative of the function. So, let's do some math acrobatics!
f(x) = x^4 - 8x^2
f'(x) = 4x^3 - 16x
f''(x) = 12x^2 - 16
To find where f''(x) is positive (indicating concave upward), we need to solve the inequality:
12x^2 - 16 > 0
After some juggling (or quadratic equation solving, if you prefer), we find:
x < -2/√3 or x > 2/√3
Now we have the intervals where f(x) is concave upward. But why stop at just giving you that information? Here's a little humor for you:
The largest open intervals where the function is concave upward are like a circus tent that goes on for miles! They stretch from negative infinity all the way to -2/√3, and then continue from 2/√3 to positive infinity. So, get ready for a wild ride full of laughter and curves! Enjoy the show! 🎪😄
To find the largest open intervals where the function is concave upward, we need to find the intervals where the second derivative is positive.
First, let's find the first derivative of the function f(x):
f'(x) = 4x^3 - 16x
Now, let's find the second derivative by taking the derivative of f'(x):
f''(x) = 12x^2 - 16
To find where f''(x) > 0, we set the second derivative greater than zero and solve for x:
12x^2 - 16 > 0
Adding 16 to both sides:
12x^2 > 16
Dividing both sides by 12:
x^2 > 16/12
Simplifying:
x^2 > 4/3
Taking the square root of both sides:
x > ± √(4/3)
So, the function f(x) is concave upward for x values greater than √(4/3) and less than -√(4/3).
Therefore, the largest open intervals where the function is concave upward are:
(-∞, -√(4/3)) and (√(4/3), +∞)
To find the largest open intervals where the function is concave upward, we need to find the intervals where the second derivative of the function, or the concavity, is positive.
First, let's find the second derivative of the function f(x). The second derivative represents the rate at which the slope of the function is changing. If the second derivative is positive, it indicates a concave upward shape.
Given f(x) = x^4 - 8x^2, we differentiate it once to find the first derivative, f'(x):
f'(x) = 4x^3 - 16x
Now, differentiate f'(x) to find the second derivative, f''(x):
f''(x) = 12x^2 - 16
To find the largest open intervals where f(x) is concave upward, we need to find the values of x for which f''(x) is positive.
Setting f''(x) > 0, we have:
12x^2 - 16 > 0
Solving this inequality, we can factor it as follows:
4(3x^2 - 4) > 0
To find the critical points where the inequality changes sign, set each factor equal to zero:
3x^2 - 4 = 0
x^2 - 4/3 = 0
(x + 2/√3)(x - 2/√3) = 0
The critical points are x = -2/√3 and x = 2/√3.
Now, we can create a number line with these critical points and test each interval to see where f''(x) is positive.
On the number line, we'll place the critical points -2/√3 and 2/√3. We will test three intervals: (-∞, -2/√3), (-2/√3, 2/√3), and (2/√3, ∞).
Testing the interval (-∞, -2/√3):
Choose a test point within this interval, for example, x = -1. Plug it into the inequality f''(x) > 0:
12(-1)^2 - 16 > 0
12 - 16 > 0
-4 > 0
Since the test point gives a negative result, f''(x) is not positive in this interval.
Testing the interval (-2/√3, 2/√3):
Choose a test point within this interval, for example, x = 0. Plug it into the inequality f''(x) > 0:
12(0)^2 - 16 > 0
0 - 16 > 0
-16 > 0
Again, the test point gives a negative result, so f''(x) is not positive in this interval either.
Testing the interval (2/√3, ∞):
Choose a test point within this interval, for example, x = 1. Plug it into the inequality f''(x) > 0:
12(1)^2 - 16 > 0
12 - 16 > 0
-4 > 0
Once again, the test point gives a negative result, meaning that f''(x) is not positive in this interval.
Therefore, we found that f''(x) is not positive in any of the tested intervals.
As a result, there are no open intervals where the function f(x) = x^4 - 8x^2 is concave upward.