Math / Algebra

Question $f(x)=\left\{\begin{array}{ll} e^{b x} & \text { for } x \leq 2 \\ 1.5 x+b & \text { for } x>2 \end{array}\right.$ et $f$ be the function defined above. For what values of $b$ is $f$ continuous at $x=2$ ? (A) 0.508 only (B) 0.647 only (C) -1.282 and 0.508
D -2.998 and 0.647

Studdy Solution

STEP 1

What is this asking? We need to find the value(s) of $b$ that make the piecewise function $f(x)$ connect smoothly at $x = 2$ . Watch out! A common mistake is to only check one side of the continuity condition.
We need to make sure the left-hand limit, the right-hand limit, and the function's value at $x = 2$ are all the same!

STEP 2

1. Define Continuity
2. Set up the Equations
3. Solve for b

STEP 3

For a function to be continuous at a point, the left-hand limit, the right-hand limit, and the function's value at that point must all be equal.
Super important!
In our case, we need to make sure everything lines up perfectly at $x = 2$ .

STEP 4

Let's look at the left-hand limit.
As $x$ approaches $\mathbf{2}$ from the left, we use the first piece of the function: $f(x) = e^{bx}$ .
So, the left-hand limit is $\lim_{x \to 2^-} f(x) = e^{b \cdot 2} = e^{2b}$ .

STEP 5

Now, the right-hand limit.
As $x$ approaches $\mathbf{2}$ from the right, we use the second piece: $f(x) = 1.5x + b$ .
Thus, the right-hand limit is $\lim_{x \to 2^+} f(x) = 1.5 \cdot 2 + b = 3 + b$ .

STEP 6

Finally, the function's value at $x = 2$ .
Since the first piece of the function is defined for $x \leq 2$ , we use $f(x) = e^{bx}$ to find $f(2) = e^{b \cdot 2} = e^{2b}$ .

STEP 7

For continuity at $x = 2$ , all three of these must be equal: $e^{2b} = 3 + b$ .
This is the key equation we need to solve!

STEP 8

This equation isn't easy to solve directly!
We'll need a bit of cleverness.
Notice that if $b = 0.508$ , we have $e^{2 \cdot 0.508} \approx e^{1.016} \approx 2.76$ , and $3 + 0.508 = 3.508$ .
Close, but not quite!

STEP 9

Let's try $b = 0.647$ .
We get $e^{2 \cdot 0.647} \approx e^{1.294} \approx 3.646$ , and $3 + 0.647 = 3.647$ .
Wow, that's incredibly close!
So, $b \approx \mathbf{0.647}$ is a very good approximation.

STEP 10

We can also see that there's another solution.
If we try a negative value for $b$ , say $b = -1.282$ , we get $e^{2 \cdot (-1.282)} \approx e^{-2.564} \approx 0.077$ , and $3 + (-1.282) = 1.718$ .
Not a match.

STEP 11

If we try $b = -2.998$ , we get $e^{2 \cdot (-2.998)} \approx e^{-5.996} \approx 0.0025$ , and $3 + (-2.998) = 0.002$ .
Again, very close!
So, $b \approx \mathbf{-2.998}$ is another good approximation.

STEP 12

The values of $b$ that make $f(x)$ continuous at $x = 2$ are approximately $b = \mathbf{0.647}$ and $b = \mathbf{-2.998}$ .
This corresponds to answer choice (D).

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Question $f(x)=\left\{\begin{array}{ll} e^{b x} & \text { for } x \leq 2 \\ 1.5 x+b & \text { for } x>2 \end{array}\right.$ et $f$ be the function defined above. For what values of $b$ is $f$ continuous at $x=2$ ? (A) 0.508 only (B) 0.647 only (C) -1.282 and 0.508
D -2.998 and 0.647

Studdy Solution

STEP 1

STEP 2

1. Define Continuity
2. Set up the Equations
3. Solve for b

STEP 3

For a function to be continuous at a point, the left-hand limit, the right-hand limit, and the function's value at that point must all be equal.
Super important!
In our case, we need to make sure everything lines up perfectly at $x = 2$ .

STEP 4

Let's look at the left-hand limit.
As $x$ approaches $\mathbf{2}$ from the left, we use the first piece of the function: $f(x) = e^{bx}$ .
So, the left-hand limit is $\lim_{x \to 2^-} f(x) = e^{b \cdot 2} = e^{2b}$ .

STEP 5

Now, the right-hand limit.
As $x$ approaches $\mathbf{2}$ from the right, we use the second piece: $f(x) = 1.5x + b$ .
Thus, the right-hand limit is $\lim_{x \to 2^+} f(x) = 1.5 \cdot 2 + b = 3 + b$ .

STEP 6

Finally, the function's value at $x = 2$ .
Since the first piece of the function is defined for $x \leq 2$ , we use $f(x) = e^{bx}$ to find $f(2) = e^{b \cdot 2} = e^{2b}$ .

STEP 7

For continuity at $x = 2$ , all three of these must be equal: $e^{2b} = 3 + b$ .
This is the key equation we need to solve!

STEP 8

This equation isn't easy to solve directly!
We'll need a bit of cleverness.
Notice that if $b = 0.508$ , we have $e^{2 \cdot 0.508} \approx e^{1.016} \approx 2.76$ , and $3 + 0.508 = 3.508$ .
Close, but not quite!

STEP 9

Let's try $b = 0.647$ .
We get $e^{2 \cdot 0.647} \approx e^{1.294} \approx 3.646$ , and $3 + 0.647 = 3.647$ .
Wow, that's incredibly close!
So, $b \approx \mathbf{0.647}$ is a very good approximation.

STEP 10

We can also see that there's another solution.
If we try a negative value for $b$ , say $b = -1.282$ , we get $e^{2 \cdot (-1.282)} \approx e^{-2.564} \approx 0.077$ , and $3 + (-1.282) = 1.718$ .
Not a match.

STEP 11

If we try $b = -2.998$ , we get $e^{2 \cdot (-2.998)} \approx e^{-5.996} \approx 0.0025$ , and $3 + (-2.998) = 0.002$ .
Again, very close!
So, $b \approx \mathbf{-2.998}$ is another good approximation.

STEP 12

The values of $b$ that make $f(x)$ continuous at $x = 2$ are approximately $b = \mathbf{0.647}$ and $b = \mathbf{-2.998}$ .
This corresponds to answer choice (D).

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Studdy Solution

STEP 1

STEP 2

1. Define Continuity2. Set up the Equations3. Solve for *b*

STEP 3

For a function to be continuous at a point, the left-hand limit, the right-hand limit, and the function's value at that point must all be equal.Super important!In our case, we need to make sure everything lines up perfectly at x=2x = 2x=2.

STEP 4

Let's look at the left-hand limit.As xxx approaches 2\mathbf{2}2 from the left, we use the first piece of the function: f(x)=ebxf(x) = e^{bx}f(x)=ebx.So, the left-hand limit is lim⁡x→2−f(x)=eb⋅2=e2b\lim_{x \to 2^-} f(x) = e^{b \cdot 2} = e^{2b}limx→2−​f(x)=eb⋅2=e2b.

STEP 5

Now, the right-hand limit.As xxx approaches 2\mathbf{2}2 from the right, we use the second piece: f(x)=1.5x+bf(x) = 1.5x + bf(x)=1.5x+b.Thus, the right-hand limit is lim⁡x→2+f(x)=1.5⋅2+b=3+b\lim_{x \to 2^+} f(x) = 1.5 \cdot 2 + b = 3 + blimx→2+​f(x)=1.5⋅2+b=3+b.

STEP 6

Finally, the function's value at x=2x = 2x=2.Since the first piece of the function is defined for x≤2x \leq 2x≤2, we use f(x)=ebxf(x) = e^{bx}f(x)=ebx to find f(2)=eb⋅2=e2bf(2) = e^{b \cdot 2} = e^{2b}f(2)=eb⋅2=e2b.

STEP 7

For continuity at x=2x = 2x=2, all three of these must be equal: e2b=3+be^{2b} = 3 + be2b=3+b.This is the **key equation** we need to solve!

STEP 8

This equation isn't easy to solve directly!We'll need a bit of cleverness.Notice that if b=0.508b = 0.508b=0.508, we have e2⋅0.508≈e1.016≈2.76e^{2 \cdot 0.508} \approx e^{1.016} \approx 2.76e2⋅0.508≈e1.016≈2.76, and 3+0.508=3.5083 + 0.508 = 3.5083+0.508=3.508.Close, but not quite!

STEP 9

Let's try b=0.647b = 0.647b=0.647.We get e2⋅0.647≈e1.294≈3.646e^{2 \cdot 0.647} \approx e^{1.294} \approx 3.646e2⋅0.647≈e1.294≈3.646, and 3+0.647=3.6473 + 0.647 = 3.6473+0.647=3.647.Wow, that's incredibly close!So, b≈0.647b \approx \mathbf{0.647}b≈0.647 is a very good approximation.

STEP 10

STEP 11

STEP 12

The values of bbb that make f(x)f(x)f(x) continuous at x=2x = 2x=2 are approximately b=0.647b = \mathbf{0.647}b=0.647 and b=−2.998b = \mathbf{-2.998}b=−2.998.This corresponds to answer choice (D).

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1. Define Continuity
2. Set up the Equations
3. Solve for b

For a function to be continuous at a point, the left-hand limit, the right-hand limit, and the function's value at that point must all be equal.
Super important!
In our case, we need to make sure everything lines up perfectly at $x = 2$ .

Let's look at the left-hand limit.
As $x$ approaches $\mathbf{2}$ from the left, we use the first piece of the function: $f(x) = e^{bx}$ .
So, the left-hand limit is $\lim_{x \to 2^-} f(x) = e^{b \cdot 2} = e^{2b}$ .

Now, the right-hand limit.
As $x$ approaches $\mathbf{2}$ from the right, we use the second piece: $f(x) = 1.5x + b$ .
Thus, the right-hand limit is $\lim_{x \to 2^+} f(x) = 1.5 \cdot 2 + b = 3 + b$ .

Finally, the function's value at $x = 2$ .
Since the first piece of the function is defined for $x \leq 2$ , we use $f(x) = e^{bx}$ to find $f(2) = e^{b \cdot 2} = e^{2b}$ .

For continuity at $x = 2$ , all three of these must be equal: $e^{2b} = 3 + b$ .
This is the key equation we need to solve!

This equation isn't easy to solve directly!
We'll need a bit of cleverness.
Notice that if $b = 0.508$ , we have $e^{2 \cdot 0.508} \approx e^{1.016} \approx 2.76$ , and $3 + 0.508 = 3.508$ .
Close, but not quite!

Let's try $b = 0.647$ .
We get $e^{2 \cdot 0.647} \approx e^{1.294} \approx 3.646$ , and $3 + 0.647 = 3.647$ .
Wow, that's incredibly close!
So, $b \approx \mathbf{0.647}$ is a very good approximation.

The values of $b$ that make $f(x)$ continuous at $x = 2$ are approximately $b = \mathbf{0.647}$ and $b = \mathbf{-2.998}$ .
This corresponds to answer choice (D).