Find the points of inflection and intervals of concavity for f(x)=3x^4-16x^3+24^2-9
f' = 12x^3-48x^2+48x
f" = 36x^2-96x+48 = 12(3x-2)(x-2)
inflection where f" = 0
concave up where f" > 0
concave down where f" < 0
To find the points of inflection and intervals of concavity for the function f(x) = 3x^4 - 16x^3 + 24x^2 - 9, we need to follow these steps:
Step 1: Find the first and second derivatives of the function f(x).
Step 2: Solve the equation f''(x) = 0 to find the points of inflection.
Step 3: Determine the concavity between the points of inflection by analyzing the sign of the second derivative.
Let's start with step 1:
Step 1: Find the first and second derivatives of f(x).
The first derivative of f(x) is obtained by differentiating each term separately:
f'(x) = d/dx (3x^4) - d/dx (16x^3) + d/dx (24x^2) - d/dx (9)
= 12x^3 - 48x^2 + 48x
The second derivative of f(x) is obtained by differentiating the first derivative:
f''(x) = d/dx (12x^3) - d/dx (48x^2) + d/dx (48x)
= 36x^2 - 96x + 48
Now let's move on to step 2:
Step 2: Solve the equation f''(x) = 0 to find the points of inflection.
To find the points of inflection, we need to solve the equation f''(x) = 0:
36x^2 - 96x + 48 = 0
We can simplify this equation by dividing all terms by 12:
3x^2 - 8x + 4 = 0
This quadratic equation can be factored as follows:
(3x - 2)(x - 2) = 0
Setting each factor equal to zero:
3x - 2 = 0 -> x = 2/3
x - 2 = 0 -> x = 2
So, the points of inflection for f(x) are x = 2/3 and x = 2.
Finally, let's move on to step 3:
Step 3: Determine the concavity between the points of inflection by analyzing the sign of the second derivative.
To determine the concavity of f(x), we need to analyze the sign of f''(x) in different intervals.
Choose a test point in each interval of interest and plug it into f''(x) to determine the sign. We have the following intervals to consider:
Interval 1: (-∞, 2/3)
Interval 2: (2/3, 2)
Interval 3: (2, ∞)
For Interval 1, let's choose x = 0 as the test point:
f''(0) = 36(0)^2 - 96(0) + 48 = 48
Since f''(0) > 0, the concavity is positive.
For Interval 2, let's choose x = 1 as the test point:
f''(1) = 36(1)^2 - 96(1) + 48 = -12
Since f''(1) < 0, the concavity is negative.
For Interval 3, let's choose x = 3 as the test point:
f''(3) = 36(3)^2 - 96(3) + 48 = 0
Since f''(3) = 0, we cannot determine the concavity.
Therefore, the points of inflection for f(x) are x = 2/3 and x = 2, and the concavity is positive on the interval (-∞, 2/3) and negative on the interval (2/3, 2). The concavity cannot be determined for the interval (2, ∞).