Question

hw 16 - first derivative test section 3.2: problem 7
(1 point)
consider the function
$f(x)=\frac{1}{x^{2}+10x + 26}$.

a) give the domain of $f$ (in interval notation).

b) find the critical numbers of $f$. (separate multiple answers by commas.)

c) determine the intervals on which $f$ is increasing and decreasing.
$f$ is increasing on
$f$ is decreasing on

d) use the first derivative test to determine whether each critical point is a relative maximum, minimum, or neither. if the
function does not have a relative maximum or minimum, type none.
relative maxima occur at $x=$. (separate multiple answers by commas.)
relative minima occur at $x=$. (separate multiple answers by commas.)

Explanation:

Step1: Find the domain of \( f(x)=\frac{1}{x^{2}+10x + 26} \)

The denominator \(x^{2}+10x + 26=(x + 5)^{2}+1\). Since \((x + 5)^{2}\geq0\) for all real \(x\), then \((x + 5)^{2}+1>0\) for all real \(x\). So the domain is \((-\infty,\infty)\)

Step2: Find the derivative of \( f(x) \)

Using the quotient rule \(y=\frac{u}{v}\), \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here \(u = 1\), \(u^\prime=0\), \(v=x^{2}+10x + 26\), \(v^\prime=2x + 10\). Then \(f^\prime(x)=\frac{0-(2x + 10)}{(x^{2}+10x + 26)^{2}}=\frac{-(2x + 10)}{(x^{2}+10x + 26)^{2}}\)

Step3: Find the critical numbers

Set \(f^\prime(x)=0\), \(\frac{-(2x + 10)}{(x^{2}+10x + 26)^{2}}=0\). Since the denominator is always positive, set \(2x+10 = 0\), \(x=- 5\)

Step4: Determine the intervals of increase and decrease

Choose test points:

• For \(x<-5\) (e.g., \(x=-6\)), \(f^\prime(-6)=\frac{-(2\times(-6)+10)}{((-6)^{2}+10\times(-6)+26)^{2}}=\frac{-(-12 + 10)}{(36-60 + 26)^{2}}=\frac{2}{(2)^{2}}>0\)
• For \(x>-5\) (e.g., \(x=-4\)), \(f^\prime(-4)=\frac{-(2\times(-4)+10)}{((-4)^{2}+10\times(-4)+26)^{2}}=\frac{-(-8 + 10)}{(16-40 + 26)^{2}}=\frac{-2}{(2)^{2}}<0\)

So \(f(x)\) is increasing on \((-\infty,-5)\) and decreasing on \((-5,\infty)\)

Step5: Use the First - Derivative Test

Since \(f(x)\) changes from increasing to decreasing at \(x =-5\), by the First - Derivative Test, \(x=-5\) is a relative maximum. There is no relative minimum.

Answer:

a) \((-\infty,\infty)\)
b) \(-5\)
c) Increasing on \((-\infty,-5)\), Decreasing on \((-5,\infty)\)
d) Relative maxima occur at \(x=-5\), Relative minima occur at \(x=\text{none}\)