Consider the function of defined by f(x)=x^3/3-4/x
a. Find the X values for the points of inflection.
b. Determine the intervals where the function of f is concave up and concave down.
f(x)=x^3/3-4/x
= (1/3)x^3 - 4x^-1
f ' (x) = x^2 + 4x^-2
f '' (x) = 2x - 8x^-3
= 2x - 8/x^3
at the point of inflection, f ''(x) = 0
2x - 8/x^3 = 0
2x^4 = 8
x^4 = 4
x = ± √2
if x = √2 , then y = (2√2)/3 - 4/√2 = (-4√2) / 3
two inflections points : appr (1.4, -1.9) and (-1.4, 1.9)
check: confirmed by
https://www.wolframalpha.com/input/?i=graph+y+%3D+x%5E3+%2F+3+-+4%2Fx+from+-5+to+5
It should be straightforward to decide where the function is concave up or concave down
once you have the points of inflection.
- otherwise, the function is concave up when f ''(x) is positive, and concave down
if f ''(x) is negative. (of course if f ''(x) = 0 , you have the points of inflection)
To find the points of inflection and determine the concavity of the function f(x), we need to find the second derivative of the given function.
a. Find the second derivative to identify the points of inflection:
1. Start with the original function f(x) = x^3/3 - 4/x.
2. Differentiate f(x) with respect to x to find the first derivative, f'(x).
- f'(x) = d/dx (x^3/3) - d/dx (4/x)
- f'(x) = x^2 - (-4/x^2) [using the power rule and derivative of 1/x]
- f'(x) = x^2 + 4/x^2
3. Differentiate f'(x) with respect to x to find the second derivative, f''(x).
- f''(x) = d/dx (x^2 + 4/x^2)
- f''(x) = 2x - (-8/x^3) [using the power rule and derivative of 1/x^2]
- f''(x) = 2x + 8/x^3
4. To find the points of inflection, we set f''(x) equal to zero and solve for x:
- 2x + 8/x^3 = 0
- 2x = -8/x^3
- Multiply both sides by x^3 to eliminate the fraction:
2x^4 = -8
- Divide both sides by 2 to isolate x^4:
x^4 = -4
- There are no real solutions for x^4 = -4, meaning there are no points of inflection in this function.
b. To determine the intervals where the function f is concave up and concave down, we analyze the sign of the second derivative f''(x):
- Concave up: When f''(x) > 0
- Concave down: When f''(x) < 0
To determine the intervals of concavity, we examine the sign changes in the second derivative f''(x):
1. Look at the critical points where f''(x) = 0 or undefined:
- In this case, f''(x) is always defined.
2. Evaluate the sign of f''(x) in different intervals:
- When x < 0, f''(x) = 2x + 8/x^3
- Substitute a negative value for x, e.g., x = -1:
- f''(-1) = (2(-1)) + (8/(-1)^3) = -2 - 8 = -10 [negative]
- f''(x) is negative (less than zero) in this interval, indicating concavity down.
- When x > 0, f''(x) = 2x + 8/x^3
- Substitute a positive value for x, e.g., x = 1:
- f''(1) = (2(1)) + (8/(1)^3) = 2 + 8 = 10 [positive]
- f''(x) is positive (greater than zero) in this interval, indicating concavity up.
- Therefore, the function f(x) is concave down for x < 0, and concave up for x > 0.
In summary, the function f(x) has no points of inflection, and it is concave down for x < 0 and concave up for x > 0.