QuestionFind constants $a$ and $b$ for function $f(x)$ to be continuous everywhere, defined as: $f(x) = \begin{cases} a + b x, & x > 2 \\ 3, & x = 2 \\ b - a x^2, & x < 2 \end{cases}$

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is given as a piecewise function with three different expressions for different ranges of $x$ . . The function is continuous everywhere if it is continuous at every point in its domain.
3. The function is continuous at a point if the limit as $x$ approaches that point from the left is equal to the limit as $x$ approaches from the right, and both are equal to the function's value at that point.

STEP 2

To ensure the function is continuous everywhere, we need to ensure it is continuous at $x=2$ . This means the limit as $x$ approaches2 from the left (using the $b-ax^2$ expression) must equal the function's value at $x=2$ (which is), and this must also equal the limit as $x$ approaches2 from the right (using the $a+bx$ expression).
So, we have two equations to solve for $a$ and $b$ $\lim{{x \to2^-}}(b-ax^2) =$ and $\lim{{x \to2^+}}(a+bx) =$

STEP 3

Let's first solve the equation for the limit as $x$ approaches2 from the left.
$\lim{{x \to2^-}}(b-ax^2) =3$

STEP 4

Substitute $x=2$ into the equation.
$b-a(2)^2 =3$

STEP 5

implify the equation.
$b-4a =3$

STEP 6

Now, let's solve the equation for the limit as $x$ approaches2 from the right.
$\lim{{x \to2^+}}(a+bx) =3$

STEP 7

Substitute $x=2$ into the equation.
$a+b(2) =3$

STEP 8

implify the equation.
$a+2b =3$

STEP 9

Now we have a system of two equations, which we can solve to find the values of $a$ and $b$ .
$\begin{cases} b-4a =3\\a+2b =3\end{cases}$

STEP 10

Multiply the second equation by2 and subtract the first equation from the result to eliminate $b$ .
$2(a+2b) - (b-4a) =2 \cdot3 -3$

STEP 11

implify the equation to find the value of $a$ .
$a+4b-b+8a =6-3$ $10a+3b =3$ $10a =3 -3b$ $a = \frac{3 -3b}{10}$

STEP 12

Substitute $a$ into the first equation from our system of equations to find $b$ .
$b -4\left(\frac{ -b}{10}\right) =$

STEP 13

implify the equation to find the value of $b$ .
$10b -12 +12b =30$ $22b =42$ $b = \frac{42}{22} =.90909$

STEP 14

Substitute $b$ back into the equation for $a$ to find the value of $a$ .
$a = \frac{3 -3 \cdot.90909}{10}$

STEP 15

Calculate the value of $a$ .
$a = \frac{3 -5.72727}{10} = -0.272727$ The values of $a$ and $b$ that make the function $f(x)$ continuous everywhere are $a = -0.272727$ and $b =.90909$ .

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QuestionFind constants $a$ and $b$ for function $f(x)$ to be continuous everywhere, defined as: $f(x) = \begin{cases} a + b x, & x > 2 \\ 3, & x = 2 \\ b - a x^2, & x < 2 \end{cases}$

Studdy Solution

STEP 1

STEP 2

STEP 3

Let's first solve the equation for the limit as $x$ approaches2 from the left.
$\lim{{x \to2^-}}(b-ax^2) =3$

STEP 4

Substitute $x=2$ into the equation.
$b-a(2)^2 =3$

STEP 5

implify the equation.
$b-4a =3$

STEP 6

Now, let's solve the equation for the limit as $x$ approaches2 from the right.
$\lim{{x \to2^+}}(a+bx) =3$

STEP 7

Substitute $x=2$ into the equation.
$a+b(2) =3$

STEP 8

implify the equation.
$a+2b =3$

STEP 9

Now we have a system of two equations, which we can solve to find the values of $a$ and $b$ .
$\begin{cases} b-4a =3\\a+2b =3\end{cases}$

STEP 10

Multiply the second equation by2 and subtract the first equation from the result to eliminate $b$ .
$2(a+2b) - (b-4a) =2 \cdot3 -3$

STEP 11

implify the equation to find the value of $a$ .
$a+4b-b+8a =6-3$ $10a+3b =3$ $10a =3 -3b$ $a = \frac{3 -3b}{10}$

STEP 12

Substitute $a$ into the first equation from our system of equations to find $b$ .
$b -4\left(\frac{ -b}{10}\right) =$

STEP 13

implify the equation to find the value of $b$ .
$10b -12 +12b =30$ $22b =42$ $b = \frac{42}{22} =.90909$

STEP 14

Substitute $b$ back into the equation for $a$ to find the value of $a$ .
$a = \frac{3 -3 \cdot.90909}{10}$

STEP 15

Calculate the value of $a$ .
$a = \frac{3 -5.72727}{10} = -0.272727$ The values of $a$ and $b$ that make the function $f(x)$ continuous everywhere are $a = -0.272727$ and $b =.90909$ .

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QuestionFind constants aaa and bbb for function f(x)f(x)f(x) to be continuous everywhere, defined as: f(x)={a+bx,x>23,x=2b−ax2,x<2 f(x) = \begin{cases} a + b x, & x > 2 \\ 3, & x = 2 \\ b - a x^2, & x < 2 \end{cases} f(x)=⎩⎨⎧​a+bx,3,b−ax2,​x>2x=2x<2​

Studdy Solution

STEP 1

STEP 2

STEP 3

Let's first solve the equation for the limit as xxx approaches2 from the left.lim⁡x→2−(b−ax2)=3\lim{{x \to2^-}}(b-ax^2) =3limx→2−(b−ax2)=3

STEP 4

Substitute x=2x=2x=2 into the equation.b−a(2)2=3b-a(2)^2 =3b−a(2)2=3

STEP 5

implify the equation.b−4a=3b-4a =3b−4a=3

STEP 6

Now, let's solve the equation for the limit as xxx approaches2 from the right.lim⁡x→2+(a+bx)=3\lim{{x \to2^+}}(a+bx) =3limx→2+(a+bx)=3

STEP 7

Substitute x=2x=2x=2 into the equation.a+b(2)=3a+b(2) =3a+b(2)=3

STEP 8

implify the equation.a+2b=3a+2b =3a+2b=3

STEP 9

Now we have a system of two equations, which we can solve to find the values of aaa and bbb.{b−4a=3a+2b=3\begin{cases} b-4a =3\\a+2b =3\end{cases}{b−4a=3a+2b=3​

STEP 10

Multiply the second equation by2 and subtract the first equation from the result to eliminate bbb.2(a+2b)−(b−4a)=2⋅3−32(a+2b) - (b-4a) =2 \cdot3 -32(a+2b)−(b−4a)=2⋅3−3

STEP 11

implify the equation to find the value of aaa.a+4b−b+8a=6−3a+4b-b+8a =6-3a+4b−b+8a=6−310a+3b=310a+3b =310a+3b=310a=3−3b10a =3 -3b10a=3−3ba=3−3b10a = \frac{3 -3b}{10}a=103−3b​

STEP 12

Substitute aaa into the first equation from our system of equations to find bbb.b−4(−b10)=b -4\left(\frac{ -b}{10}\right) =b−4(10−b​)=

STEP 13

implify the equation to find the value of bbb.10b−12+12b=3010b -12 +12b =3010b−12+12b=3022b=4222b =4222b=42b=4222=.90909b = \frac{42}{22} =.90909b=2242​=.90909

STEP 14

Substitute bbb back into the equation for aaa to find the value of aaa.a=3−3⋅.9090910a = \frac{3 -3 \cdot.90909}{10}a=103−3⋅.90909​

STEP 15

Calculate the value of aaa.a=3−5.7272710=−0.272727a = \frac{3 -5.72727}{10} = -0.272727a=103−5.72727​=−0.272727The values of aaa and bbb that make the function f(x)f(x)f(x) continuous everywhere are a=−0.272727a = -0.272727a=−0.272727 and b=.90909b =.90909b=.90909.

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QuestionFind constants $a$ and $b$ for function $f(x)$ to be continuous everywhere, defined as: $f(x) = \begin{cases} a + b x, & x > 2 \\ 3, & x = 2 \\ b - a x^2, & x < 2 \end{cases}$

Let's first solve the equation for the limit as $x$ approaches2 from the left.
$\lim{{x \to2^-}}(b-ax^2) =3$

Substitute $x=2$ into the equation.
$b-a(2)^2 =3$

implify the equation.
$b-4a =3$

Now, let's solve the equation for the limit as $x$ approaches2 from the right.
$\lim{{x \to2^+}}(a+bx) =3$

Substitute $x=2$ into the equation.
$a+b(2) =3$

implify the equation.
$a+2b =3$

Now we have a system of two equations, which we can solve to find the values of $a$ and $b$ .
$\begin{cases} b-4a =3\\a+2b =3\end{cases}$

Multiply the second equation by2 and subtract the first equation from the result to eliminate $b$ .
$2(a+2b) - (b-4a) =2 \cdot3 -3$

implify the equation to find the value of $a$ .
$a+4b-b+8a =6-3$ $10a+3b =3$ $10a =3 -3b$ $a = \frac{3 -3b}{10}$

Substitute $a$ into the first equation from our system of equations to find $b$ .
$b -4\left(\frac{ -b}{10}\right) =$

implify the equation to find the value of $b$ .
$10b -12 +12b =30$ $22b =42$ $b = \frac{42}{22} =.90909$

Substitute $b$ back into the equation for $a$ to find the value of $a$ .
$a = \frac{3 -3 \cdot.90909}{10}$

Calculate the value of $a$ .
$a = \frac{3 -5.72727}{10} = -0.272727$ The values of $a$ and $b$ that make the function $f(x)$ continuous everywhere are $a = -0.272727$ and $b =.90909$ .