[0/1 Points] DETAILS MY NOTES LARCALCET7 4.4.007. PREVIOUS ANSWERS ASK YOUR TEACHER PRACTICE ANOTHER Determine the open intervals on which the graph is concave upward or concave downward. (Enter your answers using interval notation. If an answer does not exist, enter DNE.) $$ f(x)=\frac{21}{x^{2}+3} $$

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The function is 
$$
f(x)=\frac{21}{x^2+3}.
$$

**Step 1. Write $$ f(x) $$ in a differentiable form.**

We rewrite the function as:
$$
f(x)=21(x^2+3)^{-1}.
$$

**Step 2. Compute the first derivative $$ f'(x) $$.**

Using the chain rule:
$$
f'(x)=21\cdot (-1)(x^2+3)^{-2}\cdot (2x) = -\frac{42x}{(x^2+3)^2}.
$$

**Step 3. Compute the second derivative $$ f''(x) $$.**

Differentiate $$ f'(x)=-42x(x^2+3)^{-2} $$ using the product rule, where we let:
$$
u(x)=-42x \quad \text{and} \quad v(x)=(x^2+3)^{-2}.
$$

First, compute the derivatives:
$$
u'(x)=-42,
$$
$$
v'(x) = -2(x^2+3)^{-3}\cdot (2x) = -\frac{4x}{(x^2+3)^3}.
$$

Now apply the product rule:
$$
f''(x)=u'(x)v(x)+u(x)v'(x)= -42(x^2+3)^{-2} + \left(-42x\right)\left(-\frac{4x}{(x^2+3)^3}\right).
$$

Simplify the second term:
$$
-42x\left(-\frac{4x}{(x^2+3)^3}\right)= \frac{168x^2}{(x^2+3)^3}.
$$

Thus,
$$
f''(x)= -\frac{42}{(x^2+3)^2} + \frac{168x^2}{(x^2+3)^3}.
$$

To combine the terms, express the first term with the common denominator $$(x^2+3)^3$$:
$$
-\frac{42}{(x^2+3)^2} = -\frac{42(x^2+3)}{(x^2+3)^3}.
$$

So,
$$
f''(x)= \frac{-42(x^2+3) + 168x^2}{(x^2+3)^3}.
$$

**Step 4. Simplify the numerator.**

Expand and combine like terms:
$$
-42x^2 - 126 + 168x^2 = 126x^2 - 126.
$$

Thus,
$$
f''(x)=\frac{126(x^2-1)}{(x^2+3)^3}.
$$

**Step 5. Determine the intervals of concavity.**

Since the denominator $$(x^2+3)^3$$ is always positive for all $$ x $$, the sign of $$ f''(x) $$ depends solely on the numerator $$ 126(x^2-1) $$.

Set the numerator equal to zero to find the possible inflection points:
$$
126(x^2-1)=0 \quad \Rightarrow \quad x^2-1=0 \quad \Rightarrow \quad x= \pm 1.
$$

Now examine the sign of $$ f''(x) $$ in the intervals determined by $$ x=-1 $$ and $$ x=1 $$:

- When $$ x<-1 $$: Choose $$ x=-2 $$, then $$ x^2-1=4-1=3>0 $$, so $$ f''(x)>0 $$ (concave upward).
- When $$ -11 $$: Choose $$ x=2 $$, then $$ x^2-1=4-1=3>0 $$, so $$ f''(x)>0 $$ (concave upward).

**Final Answer:**

- The graph is concave upward on the intervals $$ (-\infty, -1) $$ and $$ (1, \infty) $$.
- The graph is concave downward on the interval $$ (-1, 1) $$.
The graph of $$ f(x) = \frac{21}{x^2 + 3} $$ is concave upward on $$ (-\infty, -1) $$ and $$ (1, \infty) $$, and concave downward on $$ (-1, 1) $$.

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