Question

consider the following function.
larcalc12 3.4.009.ep
f(x)=\frac{13}{x^{2}+12}
find the first and second derivatives.
f^{prime}(x)=
f^{prime prime}(x)=
find any values of ( c ) such that ( f^{prime prime}(c)=0 ). (enter your answer as a comma-separated list. if any answer does
not exist, enter dne.)
c=
determine the open intervals on which the graph of the function is concave upward or concave downward.
(enter your answers using interval notation. if an answer does not exist, enter dn
concave upward
concave downward

Explanation:

Step1: Find the first derivative using the quotient rule

The quotient rule is \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here \(u = 13\), \(u^\prime=0\), \(v=x^{2}+12\), \(v^\prime = 2x\).

$$
LATEXBLOCK0
$$

Step2: Find the second derivative using the quotient rule

Now, for \(y = f^\prime(x)=\frac{-26x}{(x^{2}+12)^{2}}\), \(u=-26x\), \(u^\prime=-26\), \(v=(x^{2}+12)^{2}\), and using the chain - rule \(v^\prime = 2(x^{2}+12)\times2x=4x(x^{2}+12)\)

$$
LATEXBLOCK1
$$

Step3: Solve \(f^{\prime\prime}(c) = 0\)

Set \(f^{\prime\prime}(x)=0\), so \(\frac{78x^{2}-312}{(x^{2}+12)^{3}}=0\). Since the denominator \((x^{2}+12)^{3}>0\) for all real \(x\), we solve \(78x^{2}-312 = 0\)

$$
LATEXBLOCK2
$$

Step4: Determine concavity

• Test intervals:
• For \(x<-2\), let \(x=-3\). Then \(f^{\prime\prime}(-3)=\frac{78\times(-3)^{2}-312}{((-3)^{2}+12)^{3}}=\frac{702 - 312}{(9 + 12)^{3}}=\frac{390}{21^{3}}>0\)
• For \(-2
• For \(x>2\), let \(x = 3\). Then \(f^{\prime\prime}(3)=\frac{78\times3^{2}-312}{(3^{2}+12)^{3}}=\frac{702 - 312}{(9 + 12)^{3}}=\frac{390}{21^{3}}>0\)

Answer:

\(f^\prime(x)=\frac{-26x}{(x^{2}+12)^{2}}\)

\(f^{\prime\prime}(x)=\frac{78x^{2}-312}{(x^{2}+12)^{3}}\)

\(c=-2,2\)

Concave upward: \((-\infty,-2)\cup(2,\infty)\)

Concave downward: \((-2,2)\)