Math / Calculus

QuestionThe graph of $h$ is shown above and consists of four linear pieces on the interval $-6 \leq x \leq 6$ . Find a value for the constant $b$ such that the average rate of change of $h(x)$ from $x=1$ to $x=b$ equals the following values.
1. $\mathrm{AROC}=-1$
2. $\mathrm{AROC}=0$
3. AROC $=\frac{1}{5}$
1. $b=$ $\qquad$ 2. $b=$ $\qquad$ 3. $b=$ $\qquad$ $\qquad$

Studdy Solution

STEP 1

What is this asking? We need to find the value of $b$ so that the average rate of change of the function $h(x)$ between $x = 1$ and $x = b$ is equal to -1, 0, and 1/5. Watch out! Remember the average rate of change is just the slope between two points on the graph!
Don't mix it up with the instantaneous rate of change.

STEP 2

1. Find h(1)
2. Calculate b for AROC = -1
3. Calculate b for AROC = 0
4. Calculate b for AROC = 1/5

STEP 3

Alright, let's kick things off by finding $h(1)$ !
Looking at the graph, when $x = 1$ , it looks like $h(1)$ is smack-dab in the middle of the line segment connecting $(0,0)$ and $(2,2)$ .

STEP 4

Since it's a straight line, we can just find the midpoint!
The midpoint formula is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ .
Plugging in our points $(0,0)$ and $(2,2)$ , we get $\left(\frac{0+2}{2}, \frac{0+2}{2}\right) = (1,1)$ .
So, $h(1) = 1$ .
Awesome!

STEP 5

Now, let's tackle the first average rate of change, which is $-1$ .
Remember, the average rate of change between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\frac{y_2 - y_1}{x_2 - x_1}$ .

STEP 6

We want the AROC between $x = 1$ and $x = b$ to be $-1$ .
We know $h(1) = 1$ , so we have $\frac{h(b) - 1}{b - 1} = -1$ .

STEP 7

Let's look at the graph!
We can see that when $b=6$ , $h(b) = h(6) = 0$ .
So, let's test it out! $\frac{0 - 1}{6 - 1} = \frac{-1}{5}$ .
Hmm, not $-1$ .

STEP 8

Let's try $b=0$ . $h(0)=0$ , so $\frac{0-1}{0-1} = \frac{-1}{-1} = 1$ .
Nope, not $-1$ either.

STEP 9

Let's try when $h(b)=4$ , which occurs at $b=-3$ . $\frac{4-1}{-3-1} = \frac{3}{-4}$ .
Still no luck!

STEP 10

Let's try when $h(b)=2$ , which occurs at $b=2$ . $\frac{2-1}{2-1} = \frac{1}{1} = 1$ .
Getting closer, but still not $-1$ .

STEP 11

Multiplying both sides of $\frac{h(b) - 1}{b - 1} = -1$ by $b-1$ , we get $h(b) - 1 = -b + 1$ , so $h(b) = -b + 2$ .
This is the equation of a line with slope $-1$ and y-intercept $2$ .

STEP 12

The line segment from $x=0$ to $x=2$ has the equation $h(x) = x$ .
Setting this equal to $-b+2$ , we get $x=-b+2$ .
Since we're considering $x=b$ , we have $b=-b+2$ , so $2b=2$ , and $b=1$ .
But we're looking for a $b$ different from 1.

STEP 13

Let's consider the line segment from $x=-3$ to $x=0$ .
Its equation is $h(x) = -\frac{4}{3}x + 4$ .
Setting this equal to $-b+2$ , we get $-\frac{4}{3}x + 4 = -x + 2$ .
If $x=b$ , we have $-\frac{4}{3}b + 4 = -b + 2$ .
Multiplying by 3, we get $-4b + 12 = -3b + 6$ , so $6 = b$ .

STEP 14

Let's check: $\frac{h(6)-h(1)}{6-1} = \frac{0-1}{6-1} = -\frac{1}{5}$ .
That's not right.
Let's stick with $b=0$ . $\frac{h(0)-h(1)}{0-1} = \frac{0-1}{0-1} = 1$ .
So, if we want AROC = -1, let's try $b=2$ : $\frac{h(2)-h(1)}{2-1} = \frac{2-1}{2-1} = 1$ .
It appears there's no value of $b$ that gives us an AROC of $-1$ .

STEP 15

We want $\frac{h(b) - 1}{b - 1} = 0$ .
This means $h(b) - 1 = 0$ , so $h(b) = 1$ .

STEP 16

Looking at the graph, $h(b) = 1$ when $b$ is between 0 and 2.
Since the line segment between $x=0$ and $x=2$ is defined by $h(x)=x$ , then $h(b)=b=1$ .
Since we're looking for a $b$ different from 1, there's no other value where $h(b)=1$ .

STEP 17

We want $\frac{h(b) - 1}{b - 1} = \frac{1}{5}$ .

STEP 18

We already saw that when $b=6$ , $\frac{h(6)-1}{6-1} = \frac{0-1}{5} = -\frac{1}{5}$ .

STEP 19

Let's try $b=-3$ . $\frac{h(-3)-1}{-3-1} = \frac{4-1}{-4} = -\frac{3}{4}$ .

STEP 20

If we multiply both sides by $b-1$ , we get $h(b)-1 = \frac{1}{5}(b-1)$ , so $h(b) = \frac{1}{5}b + \frac{4}{5}$ .

STEP 21

Let's consider the line segment from $x=2$ to $x=6$ .
It has equation $h(x) = -\frac{1}{2}x + 3$ .
Setting this equal to $\frac{1}{5}b + \frac{4}{5}$ and letting $x=b$ , we get $-\frac{1}{2}b + 3 = \frac{1}{5}b + \frac{4}{5}$ .
Multiplying by 10, we get $-5b + 30 = 2b + 8$ , so $22 = 7b$ , and $b = \frac{22}{7}$ .

STEP 22

1. No such $b$ exists.
2. No such $b$ exists.
3. $b = \frac{22}{7}$

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Studdy Solution

STEP 1

STEP 2

1. Find h(1)2. Calculate b for AROC = -13. Calculate b for AROC = 04. Calculate b for AROC = 1/5

STEP 3

Alright, let's **kick things off** by finding h(1)h(1)h(1)!Looking at the graph, when x=1x = 1x=1, it looks like h(1)h(1)h(1) is **smack-dab in the middle** of the line segment connecting (0,0)(0,0)(0,0) and (2,2)(2,2)(2,2).

STEP 4

STEP 5

Now, let's **tackle** the first average rate of change, which is −1-1−1.Remember, the average rate of change between two points (x1,y1)(x_1, y_1)(x1​,y1​) and (x2,y2)(x_2, y_2)(x2​,y2​) is y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1}x2​−x1​y2​−y1​​.

STEP 6

We want the AROC between x=1x = 1x=1 and x=bx = bx=b to be −1-1−1.We know h(1)=1h(1) = 1h(1)=1, so we have h(b)−1b−1=−1\frac{h(b) - 1}{b - 1} = -1b−1h(b)−1​=−1.

STEP 7

Let's look at the graph!We can see that when b=6b=6b=6, h(b)=h(6)=0h(b) = h(6) = 0h(b)=h(6)=0.So, let's test it out! 0−16−1=−15\frac{0 - 1}{6 - 1} = \frac{-1}{5}6−10−1​=5−1​.Hmm, not −1-1−1.

STEP 8

Let's try b=0b=0b=0. h(0)=0h(0)=0h(0)=0, so 0−10−1=−1−1=1\frac{0-1}{0-1} = \frac{-1}{-1} = 10−10−1​=−1−1​=1.Nope, not −1-1−1 either.

STEP 9

Let's try when h(b)=4h(b)=4h(b)=4, which occurs at b=−3b=-3b=−3. 4−1−3−1=3−4\frac{4-1}{-3-1} = \frac{3}{-4}−3−14−1​=−43​.Still no luck!

STEP 10

Let's try when h(b)=2h(b)=2h(b)=2, which occurs at b=2b=2b=2. 2−12−1=11=1\frac{2-1}{2-1} = \frac{1}{1} = 12−12−1​=11​=1.Getting closer, but still not −1-1−1.

STEP 11

Multiplying both sides of h(b)−1b−1=−1\frac{h(b) - 1}{b - 1} = -1b−1h(b)−1​=−1 by b−1b-1b−1, we get h(b)−1=−b+1h(b) - 1 = -b + 1h(b)−1=−b+1, so h(b)=−b+2h(b) = -b + 2h(b)=−b+2.This is the equation of a line with slope −1-1−1 and y-intercept 222.

STEP 12

The line segment from x=0x=0x=0 to x=2x=2x=2 has the equation h(x)=xh(x) = xh(x)=x.Setting this equal to −b+2-b+2−b+2, we get x=−b+2x=-b+2x=−b+2.Since we're considering x=bx=bx=b, we have b=−b+2b=-b+2b=−b+2, so 2b=22b=22b=2, and b=1b=1b=1.But we're looking for a bbb *different* from 1.

STEP 13

STEP 14

STEP 15

We want h(b)−1b−1=0\frac{h(b) - 1}{b - 1} = 0b−1h(b)−1​=0.This means h(b)−1=0h(b) - 1 = 0h(b)−1=0, so h(b)=1h(b) = 1h(b)=1.

STEP 16

Looking at the graph, h(b)=1h(b) = 1h(b)=1 when bbb is between 0 and 2.Since the line segment between x=0x=0x=0 and x=2x=2x=2 is defined by h(x)=xh(x)=xh(x)=x, then h(b)=b=1h(b)=b=1h(b)=b=1.Since we're looking for a bbb *different* from 1, there's no other value where h(b)=1h(b)=1h(b)=1.

STEP 17

We want h(b)−1b−1=15\frac{h(b) - 1}{b - 1} = \frac{1}{5}b−1h(b)−1​=51​.

STEP 18

We already saw that when b=6b=6b=6, h(6)−16−1=0−15=−15\frac{h(6)-1}{6-1} = \frac{0-1}{5} = -\frac{1}{5}6−1h(6)−1​=50−1​=−51​.

STEP 19

Let's try b=−3b=-3b=−3. h(−3)−1−3−1=4−1−4=−34\frac{h(-3)-1}{-3-1} = \frac{4-1}{-4} = -\frac{3}{4}−3−1h(−3)−1​=−44−1​=−43​.

STEP 20

If we multiply both sides by b−1b-1b−1, we get h(b)−1=15(b−1)h(b)-1 = \frac{1}{5}(b-1)h(b)−1=51​(b−1), so h(b)=15b+45h(b) = \frac{1}{5}b + \frac{4}{5}h(b)=51​b+54​.

STEP 21

STEP 22

1. No such bbb exists.2. No such bbb exists.3. b=227b = \frac{22}{7}b=722​

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1. Find h(1)
2. Calculate b for AROC = -1
3. Calculate b for AROC = 0
4. Calculate b for AROC = 1/5

Alright, let's kick things off by finding $h(1)$ !
Looking at the graph, when $x = 1$ , it looks like $h(1)$ is smack-dab in the middle of the line segment connecting $(0,0)$ and $(2,2)$ .

Now, let's tackle the first average rate of change, which is $-1$ .
Remember, the average rate of change between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\frac{y_2 - y_1}{x_2 - x_1}$ .

We want the AROC between $x = 1$ and $x = b$ to be $-1$ .
We know $h(1) = 1$ , so we have $\frac{h(b) - 1}{b - 1} = -1$ .

Let's look at the graph!
We can see that when $b=6$ , $h(b) = h(6) = 0$ .
So, let's test it out! $\frac{0 - 1}{6 - 1} = \frac{-1}{5}$ .
Hmm, not $-1$ .

Let's try $b=0$ . $h(0)=0$ , so $\frac{0-1}{0-1} = \frac{-1}{-1} = 1$ .
Nope, not $-1$ either.

Let's try when $h(b)=4$ , which occurs at $b=-3$ . $\frac{4-1}{-3-1} = \frac{3}{-4}$ .
Still no luck!

Let's try when $h(b)=2$ , which occurs at $b=2$ . $\frac{2-1}{2-1} = \frac{1}{1} = 1$ .
Getting closer, but still not $-1$ .

Multiplying both sides of $\frac{h(b) - 1}{b - 1} = -1$ by $b-1$ , we get $h(b) - 1 = -b + 1$ , so $h(b) = -b + 2$ .
This is the equation of a line with slope $-1$ and y-intercept $2$ .

The line segment from $x=0$ to $x=2$ has the equation $h(x) = x$ .
Setting this equal to $-b+2$ , we get $x=-b+2$ .
Since we're considering $x=b$ , we have $b=-b+2$ , so $2b=2$ , and $b=1$ .
But we're looking for a $b$ different from 1.

We want $\frac{h(b) - 1}{b - 1} = 0$ .
This means $h(b) - 1 = 0$ , so $h(b) = 1$ .

Looking at the graph, $h(b) = 1$ when $b$ is between 0 and 2.
Since the line segment between $x=0$ and $x=2$ is defined by $h(x)=x$ , then $h(b)=b=1$ .
Since we're looking for a $b$ different from 1, there's no other value where $h(b)=1$ .

We want $\frac{h(b) - 1}{b - 1} = \frac{1}{5}$ .

We already saw that when $b=6$ , $\frac{h(6)-1}{6-1} = \frac{0-1}{5} = -\frac{1}{5}$ .

Let's try $b=-3$ . $\frac{h(-3)-1}{-3-1} = \frac{4-1}{-4} = -\frac{3}{4}$ .

If we multiply both sides by $b-1$ , we get $h(b)-1 = \frac{1}{5}(b-1)$ , so $h(b) = \frac{1}{5}b + \frac{4}{5}$ .

1. No such $b$ exists.
2. No such $b$ exists.
3. $b = \frac{22}{7}$