Consider the function below.
f(x)= (x^2)/((x-4)^2)
Find the interval where the function is concave up. (Enter your answer using interval notation.)
Find the intervals where the function is concave down. (Enter your answers using interval notation.)
For interval where the function is concave up i got (-2,4) but the homework system keeps saying its wrong
and for intervals where the function is concave down i got (-oo,-2)U(4,oo) but homework system keeps saying its wrong
f = x^2/(x-4)^2
f' = -8x/(x-4)^3
f" = 16(x+2)/(x-4)^4
f is concave up where f" > 0, or x > -2
concave down then, is everywhere else, or x < -2
Unfortunately, there is an asymptote at x=4, so f is concave up on (-2,4)U(4,∞). The graph is concave up on both sides of the asymptote.
The graph is at
http://www.wolframalpha.com/input/?i=x^2%2F%28x-4%29^2
To find the interval where the function is concave up, you can find the second derivative of the function f(x) and analyze its sign:
1. Find the first derivative: f'(x) = (2x(x-4)^2 - 2x^2(2(x-4))) / (x-4)^4
Simplifying, we get: f'(x) = (2x^3 - 8x^2 - 8x^3 + 64x - 4x^3 + 32x^2) / (x-4)^4
Combining like terms, we get: f'(x) = (-10x^3 + 24x^2 + 64x) / (x-4)^4
2. Find the second derivative:
f''(x) = [(-30x^2 + 48x + 64)(x-4)^4 - (-10x^3 + 24x^2 + 64x)(4(x-4)^3)] / (x-4)^8
Simplifying further, we get: f''(x) = (-30x^6 + 192x^5 - 464x^4 + 384x^3 + 256x^2 - 1536x) / (x-4)^8
3. Analyze the sign of f''(x):
To determine where f''(x) is positive (i.e., concave up), we need to find the interval where f''(x) > 0.
Since the denominator (x-4)^8 is always positive, we only need to consider the numerator. Setting the numerator greater than 0, we have:
-30x^6 + 192x^5 - 464x^4 + 384x^3 + 256x^2 - 1536x > 0
To determine the intervals where the inequality is satisfied, we can use a number line:
On the number line, we can identify 6 critical points by setting each factor in the numerator equal to zero:
-30x^6 = 0 (critical point at x = 0)
192x^5 = 0 (no critical points here)
-464x^4 = 0 (no critical points here)
384x^3 = 0 (critical point at x = 0)
256x^2 = 0 (critical point at x = 0)
-1536x = 0 (critical point at x = 0)
Considering the intervals between the critical points, we obtain the following intervals:
Interval 1: (-∞, 0)
Interval 2: (0, ∞)
Thus, the function is concave up in the interval (-∞, 0) U (0, ∞).
For the intervals where the function is concave down, you can use a similar process:
The function is concave down when f''(x) < 0.
Analyzing the sign of f''(x) by setting it less than zero:
-30x^6 + 192x^5 - 464x^4 + 384x^3 + 256x^2 - 1536x < 0
Once again, using a number line and the critical points found previously, we obtain the following intervals:
Interval 1: (0, 4)
Interval 2: (4, ∞)
So, the function is concave down in the interval (0, 4) U (4, ∞).
Please double-check your calculations and make sure you have entered the intervals in the correct format.
To determine the intervals where a function is concave up or concave down, we need to analyze the second derivative of the function.
Let's start by finding the first derivative of the given function, f(x). Using the quotient rule, the first derivative can be calculated as:
f'(x) = ((2x)(x - 4)^2 - (x^2)(2(x - 4)))/((x - 4)^4)
= (2x(x - 4)^2 - 2x^2(x - 4))/((x - 4)^4)
= (2x(x - 4)(x - 4) - 2x^2(x - 4))/((x - 4)^4)
= (2x(x - 4)(x - 4 - x))/((x - 4)^4)
= (-2x(x - 4))/(x - 4)^3
= -2x/(x - 4)
Now, to find the second derivative, we differentiate f'(x) with respect to x:
f''(x) = d/dx (-2x/(x - 4))
= ((-2)(x - 4) - (-2x)(1))/((x - 4)^2)
= (-2x + 8 + 2x)/((x - 4)^2)
= 8/((x - 4)^2)
Now, we need to determine the intervals where the function is concave up or concave down. For this, we examine the sign of the second derivative.
When the second derivative, f''(x), is positive, the function is concave up.
When the second derivative, f''(x), is negative, the function is concave down.
Since the second derivative, f''(x), is always positive (8/(x-4)^2>0 for all x), the given function f(x) is concave up for all real numbers except where the denominator is zero, which is x = 4. Hence, the function is concave up for the interval (-∞, 4) U (4, ∞).
Regarding the intervals where the function is concave down, since the function is concave up for all values except x = 4, there are no intervals where the function is concave down. Thus, the intervals where the function is concave down are an empty set, represented as Ø or {}, or alternatively, (-∞, ∞).
Therefore, the correct answers are:
Interval where the function is concave up: (-∞, 4) U (4, ∞)
Interval where the function is concave down: Ø or {} (Empty set) or (-∞, ∞)