1) For the function 𝑦(𝑥) = 𝑥^3 − 3𝑥^2 − 9𝑥 + 5
a) Find the critical points.
b) Determine if the critical points are maximum or minimum.
c) Determine the interval where the given function is increasing and/or decreasing.
d) Find any points of inflection.
My answers:
a) 𝑦(𝑥) = 𝑥^3 − 3𝑥^2 − 9𝑥 + 5
y'(d)=d/dx (𝑥^3 − 3𝑥^2 − 9𝑥 + 5)
y'(x)= 3x^2 - 3*2x - 9 +0
y'(x) = 3x^2 - 6x -9
3x^2 - 6x -9 = 0
3 (x^2 - 2x -3)=0
x^2 - 2x -3 = 0
(x+1)(x-3)=0
x=1 and x= -3 are the critical points
can someone check my answer for part A and help me for B, C and D.
Your answer for part A is correct. The critical points of the function 𝑦(𝑥) = 𝑥^3 − 3𝑥^2 − 9𝑥 + 5 are 𝑥 = 1 and 𝑥 = -3.
Now let's move on to the other parts:
b) To determine if the critical points are maximum or minimum, we need to use the second derivative test. The second derivative, denoted as 𝑦''(𝑥), can help us determine whether a critical point is a maximum or minimum.
To find the second derivative, we differentiate the first derivative:
𝑦'(𝑥) = 3𝑥^2 - 6𝑥 - 9
Differentiating again:
𝑦''(𝑥) = d/dx (3𝑥^2 - 6𝑥 - 9)
𝑦''(𝑥) = 6𝑥 - 6
Now, substitute the critical points into the second derivative, 𝑦''(𝑥), to determine their nature.
For 𝑥 = 1:
𝑦''(1) = 6(1) - 6 = 0
Since we get a zero value, the nature of this critical point is inconclusive. We need to perform further analysis.
For 𝑥 = -3:
𝑦''(-3) = 6(-3) - 6 = -24
Since we obtain a negative value, the critical point 𝑥 = -3 corresponds to a local maximum.
c) To determine the intervals where the given function is increasing or decreasing, we can use the critical points we found and test intervals between them. We can use the first derivative to do this.
From part A, we found that the critical points are 𝑥 = 1 and 𝑥 = -3. These points divide the real number line into three intervals: (-∞, -3), (-3, 1), and (1, ∞).
Now, we test a value within each interval to see if the function is increasing or decreasing. Let's test the values of -4, 0, and 2.
For 𝑥 = -4:
𝑦'(-4) = 3(-4)^2 - 6(-4) - 9 = 63
Since we get a positive value, the function is increasing in the interval (-∞, -3).
For 𝑥 = 0:
𝑦'(0) = 3(0)^2 - 6(0) - 9 = -9
Since we get a negative value, the function is decreasing in the interval (-3, 1).
For 𝑥 = 2:
𝑦'(2) = 3(2)^2 - 6(2) - 9 = 3
Since we get a positive value, the function is increasing in the interval (1, ∞).
Therefore, the function is increasing in the intervals (-∞, -3) and (1, ∞), and decreasing in the interval (-3, 1).
d) Points of inflection occur where the concavity of a function changes. To find points of inflection, we need to analyze the concavity of the function. This is determined by the second derivative of the function.
From part b, we found that 𝑦''(𝑥) = 6𝑥 - 6.
To investigate points of inflection, we set 𝑦''(𝑥) = 0 and solve for 𝑥:
6𝑥 - 6 = 0
6𝑥 = 6
𝑥 = 1
Therefore, 𝑥 = 1 is a potential point of inflection. To confirm if it is a point of inflection, we check the behavior of the concavity on either side of 𝑥 = 1.
For 𝑥 < 1:
Choose 𝑥 = 0:
𝑦''(0) = 6(0) - 6 = -6
Since we obtain a negative value, the function is concave downward to the left of 𝑥 = 1.
For 𝑥 > 1:
Choose 𝑥 = 2:
𝑦''(2) = 6(2) - 6 = 6
Since we obtain a positive value, the function is concave upward to the right of 𝑥 = 1.
Therefore, 𝑥 = 1 is a point of inflection.
In summary:
a) The critical points are 𝑥 = 1 and 𝑥 = -3.
b) The critical point 𝑥 = 1 corresponds to a point of inflection, and 𝑥 = -3 corresponds to a local maximum.
c) The function is increasing in the intervals (-∞, -3) and (1, ∞), and decreasing in the interval (-3, 1).
d) The point of inflection is 𝑥 = 1.
A looks good
points in A are max/min if y" is negative/positive
y is increasing if y' > 0
inflection where y" = 0