Consider the following function.
f(x) = e^(x)/ x^(4)
(a) Find the intervals of increase or decrease.
(b) Find the intervals of concavity.
(c) Find the point of inflection. (If an answer does not exist, enter DNE.)
(x, y) =
dy/dx = [x^4 e^x - 4 x^3 e^x]/x^8
= [ x e^x - 4 e^x ] / x^5
= 0 when x = 4
increasing if x>4
decreasing if x <4
what is sign of second derivative?
It is a pain to compute, sketch graph of slope'
http://www.wolframalpha.com/input/?i=plot+[+x+e^x+-+4+e^x+]+%2F+x^5
sure looks like inflection point at x = 4
To find the intervals of increase or decrease for the function f(x) = e^(x)/x^(4), we need to analyze the sign of its first derivative.
Step 1: Find the first derivative of f(x).
f'(x) = (e^x * x^4 - 4 * e^x * x^3) / x^8
Step 2: Simplify the first derivative expression.
f'(x) = e^x * (x - 4) / x^5
Step 3: Determine the critical points by finding where f'(x) equals zero or is undefined.
To do this, set the numerator of f'(x) equal to zero and solve:
e^x * (x - 4) = 0
This equation is satisfied when either e^x = 0 (which has no real solutions) or (x - 4) = 0.
Therefore, the only critical point is x = 4.
Step 4: Analyze the sign of f'(x) around the critical point and in other intervals.
For x < 4:
Choose a test value, e.g., x = 3.
Substitute this value into the simplified first derivative expression:
f'(3) = e^3 * (3 - 4) / 3^5
Since e^3 is positive and (3 - 4) is negative, f'(3) is negative.
For x > 4:
Choose a test value, e.g., x = 5.
Substitute this value into the simplified first derivative expression:
f'(5) = e^5 * (5 - 4) / 5^5
Since both e^5 and (5 - 4) are positive, f'(5) is positive.
Based on these calculations, we can determine the following intervals:
(a) The function f(x) is increasing on the interval (4, ∞) and decreasing on the interval (-∞, 4).
To find the intervals of concavity for the function f(x) = e^(x)/x^(4), we need to analyze the sign of its second derivative.
Step 1: Find the second derivative of f(x).
f''(x) = (e^x * (x - 4)^2 + e^x * x * 2 * (x - 4)) / x^6
Step 2: Simplify the second derivative expression.
f''(x) = e^x * (x^2 - 8x + 16 + 2x^2 - 8x) / x^6
f''(x) = e^x * (3x^2 - 16x + 16) / x^6
Step 3: Determine the critical points by finding where f''(x) equals zero or is undefined.
To do this, set the numerator of f''(x) equal to zero and solve:
3x^2 - 16x + 16 = 0
This equation has no real solutions.
Step 4: Analyze the sign of f''(x) around the critical point and in other intervals.
Choose a test value, e.g., x = 5.
Substitute this value into the simplified second derivative expression:
f''(5) = e^5 * (3 * 5^2 - 16 * 5 + 16) / 5^6
Since e^5 is positive, (3 * 5^2 - 16 * 5 + 16) is negative, and 5^6 is positive, f''(5) is negative.
Based on this calculation, we can determine the following interval:
(b) The function f(x) is concave down on the entire real number line, (-∞, ∞).
To find the point of inflection for the function f(x) = e^(x)/x^(4), we need to find where the concavity changes.
Since the function is always concave down, there is no point of inflection.
Therefore, the point of inflection for this function is DNE (does not exist).
(c) There is no point of inflection for the function f(x) = e^(x)/x^(4).
To find the intervals of increase or decrease, we need to find the first derivative of the function and determine where it is positive or negative.
(a) Find the first derivative of f(x):
f'(x) = (e^x * x^4 - 4 * e^x * x^3) / x^8
Simplifying:
f'(x) = (e^x * x^3 * (x - 4)) / x^8
Now, we want to find the critical points by setting the numerator equal to zero:
e^x = 0 or x - 4 = 0
Since e^x is never equal to zero, we only need to consider the equation x - 4 = 0, which gives us x = 4.
Next, we can create a sign chart to analyze the sign of the derivative:
(-∞) | (0) | (4) | (∞)
---------------------------------
f'(x) | + | - | + | +
From the sign chart, we can conclude that f(x) is increasing on the intervals (-∞, 0) and (4, ∞), and decreasing on the interval (0, 4).
(b) To find the intervals of concavity, we need to find the second derivative of f(x) and determine where it is positive or negative.
Find the second derivative of f(x):
f''(x) = (e^x * x^8 + 12 * e^x * x^7 - 60 * e^x * x^6) / x^16
Simplifying:
f''(x) = (e^x * x^6 * (x^2 + 12x - 60)) / x^16
Now, we want to find the critical points by setting the numerator equal to zero:
x^2 + 12x - 60 = 0
Using the quadratic formula, we can find the solutions to this equation:
x = (-12 ± √(12^2 - 4*(-60))) / 2
Simplifying:
x = (-12 ± √(144 + 240)) / 2
x = (-12 ± √384) / 2
x = (-12 ± 19.6) / 2
x ≈ -15.8 or x ≈ 3.8
Using these critical points, we can create another sign chart to analyze the sign of the second derivative:
(-∞) | (-15.8) | (3.8) | (∞)
---------------------------------------
f''(x) | + | - | + | +
From the sign chart, we can conclude that f(x) is concave up on the intervals (-∞, -15.8) and (3.8, ∞), and concave down on the interval (-15.8, 3.8).
(c) To find the point of inflection, we need to find the x-coordinate for which the concavity changes. In this case, the concavity changes from concave up to concave down at the x-coordinate x ≈ -15.8, and from concave down to concave up at the x-coordinate x ≈ 3.8.
The point of inflection is the point (x, f(x)) where the x-coordinate is the value where the concavity changes. In this case, the points of inflection are approximately (-15.8, f(-15.8)) and (3.8, f(3.8)). To find the y-values corresponding to these x-values, plug them into the original function f(x):
(-15.8, f(-15.8)) ≈ (-15.8, e^(-15.8)/(-15.8)^4)
(3.8, f(3.8)) ≈ (3.8, e^(3.8)/(3.8)^4)
Calculating the exact y-values would require using a calculator or software.