Given the function $$ f(x)=\int_{-2}^{x}\left(t^{3}+12 t^{2}+36 t\right) d t $$, determine all intervals on which $$ f $$ is concave down.

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To determine the intervals where the function $$ f(x) = \int_{-2}^{x} \left(t^{3} + 12 t^{2} + 36 t\right) dt $$ is concave down, we need to analyze the second derivative $$ f''(x) $$.

### Step 1: Find the First Derivative $$ f'(x) $$

Using the Fundamental Theorem of Calculus:

$$
f'(x) = \frac{d}{dx} \left( \int_{-2}^{x} \left(t^{3} + 12 t^{2} + 36 t\right) dt \right) = x^{3} + 12x^{2} + 36x
$$

### Step 2: Find the Second Derivative $$ f''(x) $$

Differentiate $$ f'(x) $$ with respect to $$ x $$:

$$
f''(x) = \frac{d}{dx} \left( x^{3} + 12x^{2} + 36x \right) = 3x^{2} + 24x + 36
$$

### Step 3: Determine Where $$ f''(x) < 0 $$ (Concave Down)

Set the second derivative less than zero:

$$
3x^{2} + 24x + 36 < 0
$$

Simplify by dividing all terms by 3:

$$
x^{2} + 8x + 12 < 0
$$

Factor the quadratic equation:

$$
x^{2} + 8x + 12 = (x + 2)(x + 6)
$$

So, the inequality becomes:

$$
(x + 2)(x + 6) < 0
$$

This inequality holds true when $$ x $$ is between the roots of the equation $$ x = -6 $$ and $$ x = -2 $$.

### Conclusion

The function $$ f(x) $$ is concave down on the interval where $$ f''(x) < 0 $$, which is between $$ x = -6 $$ and $$ x = -2 $$.

**Answer:** All real numbers x between –6 and –2. In other words, f  is concave down for –6 < x < –2
The function $$ f(x) $$ is concave down on the interval $$ -6 < x < -2 $$.

Given the function , determine all intervals on
which is concave down.

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Given the function , determine all intervals on which is concave down.