Consider f(x)=200+8x^3+x^4. find the intervals on which function is increasing or decreasing, the local max and minimu values, the intervals of concavity and the inflection points
Increasing where f' > 0
Decreasing where f' < 0
f'(x) = 24x^2 + 4x^3 = 4x^2(6+x)
Concave up where f'' > 0
Concave down where f'' < 0
f'' = 48x - 12x^2 = 12x(4-x)
Max where f' = 0 and f'' < 0
Min where f' = 0 and f'' > 0
To find the intervals on which a function is increasing or decreasing, you need to determine the sign of its first derivative.
To begin, let's find the first derivative of the function f(x)=200+8x^3+x^4. The first derivative f'(x) will give us information about the function's increasing or decreasing intervals.
1. Find the first derivative:
f'(x) = d/dx(200 + 8x^3 + x^4)
= 0 + 24x^2 + 4x^3
= 4x^3 + 24x^2
Now, to determine the intervals where the function is increasing or decreasing, we need to analyze the sign of f'(x).
2. Set f'(x) equal to zero and solve for x to find the critical points:
4x^3 + 24x^2 = 0
Factoring out the common term of 4x^2:
4x^2(x + 6) = 0
Setting each factor equal to zero:
4x^2 = 0 (1st factor)
x + 6 = 0 (2nd factor)
Solving each equation gives us two critical points:
x = 0 and x = -6
Now we can create a number line based on these critical points and test different intervals.
Interval 1: (-∞, -6)
Choose a test value within this interval, such as x = -10.
Plugging it into f'(x) = 4x^3 + 24x^2, we get:
f'(-10) = 4(-10)^3 + 24(-10)^2
= -4000 + 2400
= -1600
Since the result is negative, we can conclude that f(x) is decreasing in Interval 1.
Interval 2: (-6, 0)
Choose a test value within this interval, such as x = -2.
Plugging it into f'(x) = 4x^3 + 24x^2, we get:
f'(-2) = 4(-2)^3 + 24(-2)^2
= -32 + 96
= 64
Since the result is positive, we conclude that f(x) is increasing in Interval 2.
Interval 3: (0, +∞)
Choose a test value within this interval, such as x = 2.
Plugging it into f'(x) = 4x^3 + 24x^2, we get:
f'(2) = 4(2)^3 + 24(2)^2
= 32 + 96
= 128
Since the result is positive, we can conclude that f(x) is increasing in Interval 3.
Therefore, we have found that the function f(x) = 200 + 8x^3 + x^4 is decreasing in the interval (-∞, -6) and increasing in the intervals (-6, 0) and (0, +∞).
To find the local maximum and minimum values, we need to examine the points where the function changes its increasing or decreasing behavior.
1. Let's note that the function f(x) = 200 + 8x^3 + x^4 is a continuous function.
2. We can determine the local minima and maxima by finding the critical points.
Considering the critical points we previously found, x = 0 and x = -6:
For x = 0:
f(0) = 200 + 8(0)^3 + (0)^4
= 200
For x = -6:
f(-6) = 200 + 8(-6)^3 + (-6)^4
= 200 - 1728 + 1296
= -232
Therefore, the function has a local maximum at (-6, -232) and a local minimum at (0, 200).
Now let's find the intervals of concavity and the inflection points. This requires analyzing the sign of the function's second derivative.
3. Find the second derivative:
f''(x) = d/dx(4x^3 + 24x^2)
= 12x^2 + 48x
4. Set f''(x) equal to zero and solve for x to find the possible inflection points:
12x^2 + 48x = 0
Factoring out the common term of 12x:
12x(x + 4) = 0
Setting each factor equal to zero:
12x = 0 (1st factor)
x + 4 = 0 (2nd factor)
Solving each equation gives us two critical points:
x = 0 and x = -4
Now, we can create a number line based on these critical points and test different intervals.
Interval 1: (-∞, -4)
Choose a test value within this interval, such as x = -5.
Plugging it into f''(x) = 12x^2 + 48x, we get:
f''(-5) = 12(-5)^2 + 48(-5)
= 300 - 240
= 60
Interval 2: (-4, 0)
Choose a test value within this interval, such as x = -2.
Plugging it into f''(x) = 12x^2 + 48x, we get:
f''(-2) = 12(-2)^2 + 48(-2)
= 48 - 96
= -48
Interval 3: (0, +∞)
Choose a test value within this interval, such as x = 2.
Plugging it into f''(x) = 12x^2 + 48x, we get:
f''(2) = 12(2)^2 + 48(2)
= 48 + 96
= 144
From our analysis, we can conclude that the function f(x) changes concavity at x = -4.
Thus, x = -4 is the inflection point of the function f(x) = 200 + 8x^3 + x^4.
To summarize our findings:
- The function is decreasing in the interval (-∞, -6).
- The function is increasing in the intervals (-6, 0) and (0, +∞).
- The local minimum value is (0, 200).
- The local maximum value is (-6, -232).
- The concavity changes at x = -4, which is the inflection point.
Note: It's always important to double-check the calculations and ensure the accuracy of the results obtained.