Find the points of inflection of the graph of the function. (If an answer does not exist, enter DNE.)

f(x) = 1/3x^4 + 4x^3

for some reason I am not getting the right answer; I got the graph on my calculator but it seems weird.

4/3x^3 + 12x^2 = f'(x)

4/3x^3 + 12x^2 =0
multiply by 3 to clear fractions
4x^3 + 36x^2 =0
factor out 4x^2
(4x^2)(x+9)=0
x = 0, x =-9 are critical numbers
f'(-10)= -,f'(-8) = +, f'(1) = +
(-9,-729) is an inflection, actually a minimum point

Thanks for the help but I am still confused because it's asking for

smaller x-value
larger x-value

so what would that be?

To find the points of inflection of a function, we need to follow these steps:

1. Find the second derivative of the function.
2. Set the second derivative equal to zero and solve for the values of x.
3. Determine the intervals on the x-axis where the concavity of the function changes.
4. Test a point within each interval to determine the sign of the second derivative and classify the concavity.
5. The points where the concavity changes from concave up to concave down or vice versa are the points of inflection.

Let's apply these steps to the given function f(x) = (1/3)x^4 + 4x^3:

Step 1: Find the second derivative
To find the second derivative, we need to differentiate the function twice.

f(x) = (1/3)x^4 + 4x^3
f'(x) = (4/3)x^3 + 12x^2 (First derivative)
f''(x) = 4x^2 + 24x (Second derivative)

Step 2: Set the second derivative equal to zero and solve for x:
4x^2 + 24x = 0
4x(x + 6) = 0

Setting each factor equal to zero:
4x = 0 or x + 6 = 0

Solving for x:
x = 0 or x = -6

Step 3: Determine the intervals where concavity changes:
To determine the intervals of concavity, we need to examine the behavior of the second derivative on different intervals.

To simplify this process, we can create a sign chart by selecting test points in the intervals (-∞, -6), (-6, 0), and (0, +∞). Let's choose x = -7, -1, and 1.

Using these test values, we can determine the sign of f''(x) within each interval:
For x = -7: f''(-7) = 4(-7)^2 + 24(-7) = 196 - 168 = 28 (Positive)
For x = -1: f''(-1) = 4(-1)^2 + 24(-1) = 4 - 24 = -20 (Negative)
For x = 1: f''(1) = 4(1)^2 + 24(1) = 4 + 24 = 28 (Positive)

From this, we can see that the concavity changes from concave up to concave down at x = -6 and from concave down to concave up at x = 0.

Step 4: Test a point within each interval to determine the sign of the second derivative:
We have already done this step in the previous step while determining the intervals of concavity.

Step 5: Identify the points of inflection:
The points of inflection are the points where the concavity changes. Therefore, from the previous steps, we have two points of inflection: (-6, f(-6)) and (0, f(0)).

Now that we have followed the necessary steps, we can conclude that the points of inflection for the function f(x) = (1/3)x^4 + 4x^3 are (-6, f(-6)) and (0, f(0)). If your graph on the calculator seems weird or doesn't match these inflection points, it's a good idea to double-check your calculations or consult further resources if needed.