5. (0/1 Points) DETAILS MY NOTES SCALCET9 4,1,065, PREVIOUS ANSWER Find the absolute maximum and absolute minimum values of on the given interval, $$ f(x)=\ln \left(x^{2}+5 x+14\right), \quad\{-3,1\} $$ absolute minimum value $$ \square $$ absolute maximum value $$ \square $$ Need Help? Merdil

Question

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**Step 1. Find the derivative**

We are given
$$
f(x)=\ln\left(x^2+5x+14\right).
$$
Let 
$$
g(x)=x^2+5x+14,
$$
so that
$$
f(x)=\ln\left(g(x)\right).
$$
Using the chain rule, the derivative is
$$
f'(x)=\frac{g'(x)}{g(x)}.
$$
We differentiate $$g(x)$$:
$$
g'(x)=2x+5.
$$
Thus,
$$
f'(x)=\frac{2x+5}{x^2+5x+14}.
$$

---

**Step 2. Find the critical points**

Critical points occur when $$f'(x)=0$$ or when $$f'(x)$$ is undefined. Since the denominator $$x^2+5x+14$$ is always positive (its discriminant is negative), we only need to solve
$$
2x+5=0.
$$
Solving,
$$
x=-\frac{5}{2}.
$$
We check that $$x=-\frac{5}{2}$$ is in the interval $$[-3,1]$$.

---

**Step 3. Evaluate $$f(x)$$ at the endpoints and the critical point**

1. At $$x=-3$$:
   $$
   f(-3)=\ln\left((-3)^2+5(-3)+14\right)=\ln\left(9-15+14\right)=\ln(8).
   $$

2. At $$x=-\frac{5}{2}$$:
   $$
   f\left(-\frac{5}{2}\right)=\ln\left[\left(-\frac{5}{2}\right)^2+5\left(-\frac{5}{2}\right)+14\right].
   $$
   Compute the expression inside the logarithm:
   $$
   \left(-\frac{5}{2}\right)^2=\frac{25}{4}, \quad 5\left(-\frac{5}{2}\right)=-\frac{25}{2}.
   $$
   So,
   $$
   \frac{25}{4}-\frac{25}{2}+14=\frac{25}{4}-\frac{50}{4}+\frac{56}{4}=\frac{31}{4}.
   $$
   Thus,
   $$
   f\left(-\frac{5}{2}\right)=\ln\left(\frac{31}{4}\right).
   $$

3. At $$x=1$$:
   $$
   f(1)=\ln\left(1^2+5(1)+14\right)=\ln\left(1+5+14\right)=\ln(20).
   $$

---

**Step 4. Determine the absolute maximum and minimum**

Comparing the values:
- $$f(-3)=\ln(8)$$,
- $$f\left(-\frac{5}{2}\right)=\ln\left(\frac{31}{4}\right)$$,
- $$f(1)=\ln(20)$$.

Note that
$$
\ln(8)=\ln\left(\frac{32}{4}\right),
$$
and since $$\frac{31}{4} < \frac{32}{4}$$, it follows that
$$
\ln\left(\frac{31}{4}\right) < \ln(8).
$$
Also, $$\ln(20)$$ is the largest.

Thus, the absolute minimum value is
$$
\ln\left(\frac{31}{4}\right)
$$
(at $$x=-\frac{5}{2}$$), and the absolute maximum value is
$$
\ln(20)
$$
(at $$x=1$$).

---

**Final Answer**

Absolute minimum value: $$\ln\left(\frac{31}{4}\right)$$  
Absolute maximum value: $$\ln(20)$$
Absolute minimum value: $$ \ln\left(\frac{31}{4}\right) $$ Absolute maximum value: $$ \ln(20) $$

(0/1 Points)
DETAILS
MY NOTES
SCALCET9 4,1,065,
PREVIOUS ANSWER

Find the absolute maximum and absolute minimum values of on the given interval,

absolute minimum value absolute maximum value
Need Help?
Merdil

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(0/1 Points) DETAILS MY NOTES SCALCET9 4,1,065, PREVIOUS ANSWER Find the absolute maximum and absolute minimum values of on the given interval, absolute minimum value absolute maximum value Need Help? Merdil