Find intervals of increase/decrease and classify all critical points of
(a)g(x)=cos(x/(x2 +9))
(b) f(x)=x5/3 −x2/3
can't make out a)
if x5/3 is supposed to mean x^(5/3)
use the notation I just used.
(a) g(x)=cos(x/(x^2 +9))
(b) f(x)=x^(5/3) −x^(2/3)
To find the intervals of increase and decrease and classify the critical points of a function, we can follow these steps:
Step 1: Find the derivative of the function.
Step 2: Set the derivative equal to zero and solve for x to determine the critical points.
Step 3: Determine the intervals where the derivative is positive (increasing) and negative (decreasing).
Step 4: Classify each critical point as a local maximum, local minimum, or neither by using the first or second derivative test.
Now let's apply these steps to the given functions:
(a) g(x) = cos(x/(x^2 + 9))
Step 1: Find the derivative of g(x).
g'(x) = -(sin(x/(x^2 + 9)) * (x^2 + 9) - cos(x/(x^2 + 9)) * 2x) / (x^2 + 9)^2
Step 2: Set the derivative equal to zero and solve for x.
To find the critical points, we set g'(x) = 0 and solve for x. However, the trigonometric function involved makes it difficult to find an explicit solution. Therefore, we will use a numerical method, such as graphing or using a calculator, to approximate the critical points.
Step 3: Determine the intervals of increase and decrease.
To identify the intervals of increase and decrease, we need to examine the sign of the derivative function. We can do this by analyzing the critical points and also testing points in each interval.
Step 4: Classify the critical points.
To classify the critical points as a local maximum, local minimum, or neither, we can use the first or second derivative test. However, since we are unable to determine the critical points analytically, we do not currently have enough information to classify them.
(b) f(x) = x^(5/3) − x^(2/3)
Step 1: Find the derivative of f(x).
f'(x) = (5/3)x^(2/3) − (2/3)x^(-1/3)
Step 2: Set the derivative equal to zero and solve for x.
To find the critical points, we set f'(x) = 0 and solve for x.
(5/3)x^(2/3) − (2/3)x^(-1/3) = 0
Multiplying both sides by 3, we get:
5x^(2/3) - 2x^(-1/3) = 0
Raising both sides to the power of 3 to get rid of the fractional exponents, we have:
5^3 * x^2 - 2^3 * x^(-1) = 0
125x^2 - 8 = 0
Solving for x, we find two critical points:
x = sqrt(8/125)
x ≈ ±0.267
Step 3: Determine the intervals of increase and decrease.
To determine the intervals of increase and decrease, we need to analyze the sign of the derivative function. We can test points in each interval to help us.
If we choose x = 0 as a test point, we find f'(0) = -(2/3)x^(-1/3) = -∞. Since the derivative is negative in the interval (-∞, 0), this interval is decreasing.
If we choose x = 1 as a test point, we find f'(1) = (5/3) - (2/3) = 1. Since the derivative is positive in the interval (0, ∞), this interval is increasing.
Step 4: Classify the critical points.
To classify the critical points, we can use either the first or second derivative test. Taking the second derivative, we can determine the concavity of the function.
f''(x) = (10/9)x^(-1/3) + (2/9)x^(-4/3)
Substituting the critical points into the second derivative:
f''(sqrt(8/125)) = (5/9)(1/sqrt(8/125)) + (1/9)(1/(8/125)) ≈ 1.92
f''(-sqrt(8/125)) = (5/9)(1/sqrt(8/125)) + (1/9)(1/(8/125)) ≈ 1.92
Since both second derivatives are positive, the critical points x ≈ ±0.267 are classified as local minima.
In summary:
(a) For the function g(x) = cos(x/(x^2 + 9)), we were unable to determine the critical points analytically due to the trigonometric function involved.
(b) For the function f(x) = x^(5/3) − x^(2/3), we found two critical points at x ≈ ±0.267. The interval (-∞, 0) is decreasing, and the interval (0, ∞) is increasing. The critical points x ≈ ±0.267 are classified as local minima.