For the function f(x) = 3x^3 - 4x^2 + 1, what interval indicates where the function is concave down?

To determine where the function f(x) = 3x^3 - 4x^2 + 1 is concave down, we need to analyze the second derivative of the function.

First, we find the first derivative, f'(x):

f'(x) = 9x^2 - 8.

Next, we compute the second derivative, f''(x):

f''(x) = 18x.

The concavity of a function is indicated by the sign of the second derivative. Specifically, if f''(x) is less than zero, the function is concave down.

Setting the second derivative to zero allows us to find the inflection point:

18x = 0 → x = 0.

Now, we analyze the sign of f''(x) on either side of x = 0.

• For x < 0, f''(x) = 18x < 0, indicating that the function is concave down on the interval (-∞, 0).

• For x > 0, f''(x) = 18x > 0, indicating that the function is concave up on the interval (0, ∞).

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