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TANAPMATH7 10.1.029.EP. PREVIOUS ANSWERS
Consider the following function. $g(t)=\frac{9 t}{t^{2}+4}$
Find the derivative of the function. $g^{\prime}(t)=\frac{-9 t^{2}+36}{\left(t^{2}+4\right)^{2}} \quad$ Nice job. Find all the values of $t$ for which $g^{\prime}(t)=0$ or $g^{\prime}(t)$ is discontinuous. (Enter your answers as a comma-separated list.) $t=$ $\square$ Please remove the grouping symbols from around your list. Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using interval notation. If the answer cannot be expressed as an interval increasing $\quad(-2,2)$ Terrific! decreasing $(-\infty,-2) \cup(2, \infty)$ - Good job. Need Help? Read II

Studdy Solution

STEP 1

What is this asking? We're given a function $g(t)$ and we need to find its derivative, $g'(t)$ , figure out where $g'(t)$ is zero or discontinuous, and then determine where $g(t)$ is increasing and decreasing. Watch out! Don't forget to check for discontinuities in $g'(t)$ , not just where it's zero!
Also, make sure to correctly use interval notation for increasing and decreasing intervals.

STEP 2

1. Find the derivative
2. Find where the derivative is zero or undefined
3. Determine increasing and decreasing intervals

STEP 3

We'll use the quotient rule, which says if $g(t) = \frac{u(t)}{v(t)}$ , then $g'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}$ .
Here, $u(t) = 9t$ and $v(t) = t^2 + 4$ .

STEP 4

The derivative of $u(t)$ is $u'(t) = 9$ .
The derivative of $v(t)$ is $v'(t) = 2t$ .

STEP 5

Plugging everything into the quotient rule formula, we get: $g'(t) = \frac{(9)(t^2 + 4) - (9t)(2t)}{(t^2 + 4)^2}$

STEP 6

$g'(t) = \frac{9t^2 + 36 - 18t^2}{(t^2 + 4)^2} = \frac{-9t^2 + 36}{(t^2 + 4)^2}$

STEP 7

We set the numerator equal to zero: $-9t^2 + 36 = 0$ .
Adding $9t^2$ to both sides gives $36 = 9t^2$ .
Dividing both sides by 9 gives $4 = t^2$ .
Taking the square root of both sides gives $t = \pm 2$ .
So, $g'(t) = 0$ at $t = -2$ and $t = 2$ .

STEP 8

$g'(t)$ is undefined when the denominator is zero.
However, $t^2 + 4$ is always positive, so $(t^2 + 4)^2$ is never zero.
Thus, $g'(t)$ is defined for all real numbers.

STEP 9

Our critical points are $t = -2$ and $t = 2$ .
We'll test the intervals $(-\infty, -2)$ , $(-2, 2)$ , and $(2, \infty)$ .

STEP 10

$g'(-3) = \frac{-9(-3)^2 + 36}{((-3)^2 + 4)^2} = \frac{-81 + 36}{13^2} = \frac{-45}{169} < 0$ .
So, $g(t)$ is decreasing on $(-\infty, -2)$ .

STEP 11

$g'(0) = \frac{-9(0)^2 + 36}{(0^2 + 4)^2} = \frac{36}{16} = \frac{9}{4} > 0$ .
So, $g(t)$ is increasing on $(-2, 2)$ .

STEP 12

$g'(3) = \frac{-9(3)^2 + 36}{(3^2 + 4)^2} = \frac{-81 + 36}{13^2} = \frac{-45}{169} < 0$ .
So, $g(t)$ is decreasing on $(2, \infty)$ .

STEP 13

$g'(t) = \frac{-9t^2 + 36}{(t^2 + 4)^2}$ . $g'(t) = 0$ at $t = -2, 2$ . $g'(t)$ is never discontinuous. $g(t)$ is increasing on $(-2, 2)$ and decreasing on $(-\infty, -2) \cup (2, \infty)$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the derivative
2. Find where the derivative is zero or undefined
3. Determine increasing and decreasing intervals

STEP 3

We'll use the quotient rule, which says if $g(t) = \frac{u(t)}{v(t)}$ , then $g'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}$ .
Here, $u(t) = 9t$ and $v(t) = t^2 + 4$ .

STEP 4

The derivative of $u(t)$ is $u'(t) = 9$ .
The derivative of $v(t)$ is $v'(t) = 2t$ .

STEP 5

Plugging everything into the quotient rule formula, we get: $g'(t) = \frac{(9)(t^2 + 4) - (9t)(2t)}{(t^2 + 4)^2}$

STEP 6

$g'(t) = \frac{9t^2 + 36 - 18t^2}{(t^2 + 4)^2} = \frac{-9t^2 + 36}{(t^2 + 4)^2}$

STEP 7

We set the numerator equal to zero: $-9t^2 + 36 = 0$ .
Adding $9t^2$ to both sides gives $36 = 9t^2$ .
Dividing both sides by 9 gives $4 = t^2$ .
Taking the square root of both sides gives $t = \pm 2$ .
So, $g'(t) = 0$ at $t = -2$ and $t = 2$ .

STEP 8

$g'(t)$ is undefined when the denominator is zero.
However, $t^2 + 4$ is always positive, so $(t^2 + 4)^2$ is never zero.
Thus, $g'(t)$ is defined for all real numbers.

STEP 9

Our critical points are $t = -2$ and $t = 2$ .
We'll test the intervals $(-\infty, -2)$ , $(-2, 2)$ , and $(2, \infty)$ .

STEP 10

$g'(-3) = \frac{-9(-3)^2 + 36}{((-3)^2 + 4)^2} = \frac{-81 + 36}{13^2} = \frac{-45}{169} < 0$ .
So, $g(t)$ is decreasing on $(-\infty, -2)$ .

STEP 11

$g'(0) = \frac{-9(0)^2 + 36}{(0^2 + 4)^2} = \frac{36}{16} = \frac{9}{4} > 0$ .
So, $g(t)$ is increasing on $(-2, 2)$ .

STEP 12

$g'(3) = \frac{-9(3)^2 + 36}{(3^2 + 4)^2} = \frac{-81 + 36}{13^2} = \frac{-45}{169} < 0$ .
So, $g(t)$ is decreasing on $(2, \infty)$ .

STEP 13

$g'(t) = \frac{-9t^2 + 36}{(t^2 + 4)^2}$ . $g'(t) = 0$ at $t = -2, 2$ . $g'(t)$ is never discontinuous. $g(t)$ is increasing on $(-2, 2)$ and decreasing on $(-\infty, -2) \cup (2, \infty)$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the derivative2. Find where the derivative is zero or undefined3. Determine increasing and decreasing intervals

STEP 3

STEP 4

The derivative of u(t)u(t)u(t) is u′(t)=9u'(t) = 9u′(t)=9.The derivative of v(t)v(t)v(t) is v′(t)=2tv'(t) = 2tv′(t)=2t.

STEP 5

Plugging everything into the quotient rule formula, we get: g′(t)=(9)(t2+4)−(9t)(2t)(t2+4)2g'(t) = \frac{(9)(t^2 + 4) - (9t)(2t)}{(t^2 + 4)^2}g′(t)=(t2+4)2(9)(t2+4)−(9t)(2t)​

STEP 6

g′(t)=9t2+36−18t2(t2+4)2=−9t2+36(t2+4)2g'(t) = \frac{9t^2 + 36 - 18t^2}{(t^2 + 4)^2} = \frac{-9t^2 + 36}{(t^2 + 4)^2}g′(t)=(t2+4)29t2+36−18t2​=(t2+4)2−9t2+36​

STEP 7

STEP 8

g′(t)g'(t)g′(t) is undefined when the denominator is zero.However, t2+4t^2 + 4t2+4 is always positive, so (t2+4)2(t^2 + 4)^2(t2+4)2 is never zero.Thus, g′(t)g'(t)g′(t) is defined for all real numbers.

STEP 9

Our critical points are t=−2t = -2t=−2 and t=2t = 2t=2.We'll test the intervals (−∞,−2)(-\infty, -2)(−∞,−2), (−2,2)(-2, 2)(−2,2), and (2,∞)(2, \infty)(2,∞).

STEP 10

g′(−3)=−9(−3)2+36((−3)2+4)2=−81+36132=−45169<0g'(-3) = \frac{-9(-3)^2 + 36}{((-3)^2 + 4)^2} = \frac{-81 + 36}{13^2} = \frac{-45}{169} < 0g′(−3)=((−3)2+4)2−9(−3)2+36​=132−81+36​=169−45​<0.So, g(t)g(t)g(t) is decreasing on (−∞,−2)(-\infty, -2)(−∞,−2).

STEP 11

g′(0)=−9(0)2+36(02+4)2=3616=94>0g'(0) = \frac{-9(0)^2 + 36}{(0^2 + 4)^2} = \frac{36}{16} = \frac{9}{4} > 0g′(0)=(02+4)2−9(0)2+36​=1636​=49​>0.So, g(t)g(t)g(t) is increasing on (−2,2)(-2, 2)(−2,2).

STEP 12

g′(3)=−9(3)2+36(32+4)2=−81+36132=−45169<0g'(3) = \frac{-9(3)^2 + 36}{(3^2 + 4)^2} = \frac{-81 + 36}{13^2} = \frac{-45}{169} < 0g′(3)=(32+4)2−9(3)2+36​=132−81+36​=169−45​<0.So, g(t)g(t)g(t) is decreasing on (2,∞)(2, \infty)(2,∞).

STEP 13

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1. Find the derivative
2. Find where the derivative is zero or undefined
3. Determine increasing and decreasing intervals

The derivative of $u(t)$ is $u'(t) = 9$ .
The derivative of $v(t)$ is $v'(t) = 2t$ .

Plugging everything into the quotient rule formula, we get: $g'(t) = \frac{(9)(t^2 + 4) - (9t)(2t)}{(t^2 + 4)^2}$

$g'(t) = \frac{9t^2 + 36 - 18t^2}{(t^2 + 4)^2} = \frac{-9t^2 + 36}{(t^2 + 4)^2}$

$g'(t)$ is undefined when the denominator is zero.
However, $t^2 + 4$ is always positive, so $(t^2 + 4)^2$ is never zero.
Thus, $g'(t)$ is defined for all real numbers.

Our critical points are $t = -2$ and $t = 2$ .
We'll test the intervals $(-\infty, -2)$ , $(-2, 2)$ , and $(2, \infty)$ .

$g'(-3) = \frac{-9(-3)^2 + 36}{((-3)^2 + 4)^2} = \frac{-81 + 36}{13^2} = \frac{-45}{169} < 0$ .
So, $g(t)$ is decreasing on $(-\infty, -2)$ .

$g'(0) = \frac{-9(0)^2 + 36}{(0^2 + 4)^2} = \frac{36}{16} = \frac{9}{4} > 0$ .
So, $g(t)$ is increasing on $(-2, 2)$ .

$g'(3) = \frac{-9(3)^2 + 36}{(3^2 + 4)^2} = \frac{-81 + 36}{13^2} = \frac{-45}{169} < 0$ .
So, $g(t)$ is decreasing on $(2, \infty)$ .