Question

Let a and b represent real numbers and let f(x) be the piecewise function defined below
F(x)

Answer 1

lim​f(x)=9 using the values of a=2 and b=3
x→3

Explanation:

To find the values of aa and bb that make the piecewise function continuous on the entire real number line, we need to ensure that the function is continuous at the points where the pieces meet. In this case, the function has three pieces, so we need to check the continuity at two points: x=2 and x=5.

To check the continuity at x=2, we need to make sure that the left-hand limit, right-hand limit, and the value of the function at x=2 are all equal. The left-hand limit is the value of the function as xx approaches 22 from the left, which is given by the first piece of the function:

lim​f(x)=a(2)2+2−b=4a+2−b
x→2−
The right-hand limit is the value of the function as xx approaches 22 from the right, which is given by the second piece of the function:

lim⁡f(x)=a(2)+b=2a+b
x→2+

The value of the function at x=2x=2 is given by the first piece of the function:

f(2)=a(2)^2+2−b=4a+2−b

To check the continuity at x=5, we need to make sure that the left-hand limit, right-hand limit, and the value of the function at x=5x=5 are all equal. The left-hand limit is the value of the function as xx approaches 55 from the left, which is given by the second piece of the function:
lim​f(x)=a(5)+b=5a+b
x→5−

The right-hand limit is the value of the function as xx approaches 55 from the right, which is given by the third piece of the function:
lim​f(x)=2a(5)−7=10a−7
x→5+

To make the function continuous at x=2, we set the left-hand limit equal to the right-hand limit and solve for a and b:
4a+2−b=2a+b

Simplifying this equation, we get:
2a+2=2b

a+1=b

To make the function continuous at x=5x=5, we set the left-hand limit equal to the right-hand limit and solve for a and b:
5a+b=10a−7

Simplifying this equation, we get:
7=5a−b

Substituting a+1 for b in the second equation, we get:
7=5a−(a+1)

Simplifying this equation, we get:
7=4a−1

8=4a

a=2

Substituting a=2 into the first equation, we get:
2+1=b

b=3

Therefore, the values of a and b that make the piecewise function continuous on the entire real number line are a=2 and b=3.

To evaluate limf(x)limx→3​f(x)

⁡x→3 using the values of a and b we found, we need to consider the three pieces of the function:
For x<2, we use the first piece of the function:

f(x)=ax^2+x−b=2x^2+x−3

For 2<x<5, we use the second piece of the function:

f(x)=ax+b=2x+3

For x>5, we use the third piece of the function:

f(x)=2ax−7=4x−7

Since 3 is between 2 and 5, we use the second piece of the function to evaluate the limit:

limf(x)=lim⁡x→3(2x+3)=2(3)+3=9

⁡x→3

Therefore, lim⁡f(x)=9limx→3​f(x)=9

x→3 using the values of a=2 and b=3 we found in part (a).

Answer 2

`\textsf{a)}\quad a=2,\;\;b=3`

`\textsf{b)} \quad \displaystyle\lim_(x \to 3)f(x)= 9`

Explanation:

Given piecewise function:

`f(x)=\begin{cases}ax^2+x-b \;\;&x\leq 2\\ax+b&2 < x < 5\\2ax-7&x\geq 5\end{cases}`

To ensure that the function f(x) is continuous on the entire real number line, we need to make sure that it is continuous at each of the points where the different cases meet, which in this case are x = 2 and x = 5.

The definition of continuity at a point x = c is that the limit of the function as x approaches c from the left should be equal to the limit of the function as x approaches c from the right, and both of these should be equal to the value of the function at x = c.

`\boxed{\begin{array}{c}\textsf{If\;$f(x)$\;is\;continuous\;at\;$x=c$\;then:}\\\\\displaystyle \lim_(x \to c^(-))f(x)=f(c)\qquad\lim_(x \to c^(+))f(x)=f(c)\qquad \lim_(x \to c)f(x)=f(c)\end{array}}`

Find the limits and values of f(x) at x = 2 and x = 5 to determine the values of a and b.

At x = 2

`\textsf{Left\;limit:}\quad \displaystyle \lim_(x \to 2^(-)) a(2)^2+(2)-b=4a-b+2`

`\textsf{Right\;limit:}\quad \displaystyle \lim_(x \to 2^(+)) a(2)+b=2a+b`

`\textsf{Value of the function at $x=2$:}\quad f(2) =a(2)^2+(2)-b=4a-b+2`

According to the definition of continuity, these three values should be equal, so we have the equation:

`4a -b+2 = 2a + b`

At x = 5

`\textsf{Left\;limit:}\quad \displaystyle \lim_(x \to 5^(-)) a(5)+b=5a+b`

`\textsf{Right\;limit:}\quad \displaystyle \lim_(x \to 5^(+)) 2a(5)-7=10a-7`

`\textsf{Value of the function at $x=5$:}\quad f(5) =2a(5)-7=10a-7`

Again, according to the definition of continuity, these three values should be equal, so we have the equation:

`5a+b=10a-7`

Therefore, we have a system of linear equations:

`\begin{cases} 4a -b+2 = 2a + b\\5a+b=10a-7\end{cases}`

Solve this system of equations to find the values of a and b.

Rearrange the first equation to isolate a:

`\begin{aligned}4a -b+2 &= 2a + b\\2a&=2b-2\\a&=b-1\end{aligned}`

Substitute the expression for a into the second equation and solve for b:

`\begin{aligned}5(b-1)+b&=10(b-1)-7\\5b-5+b&=10b-10-7\\6b-5&=10b-17\\12&=4b\\b&=3\end{aligned}`

Substitute the found value of b into the expression for a and solve for a:

`\begin{aligned}a&=3-1\\a&=2\end{aligned}`

So, the values of a and b that make the function f(x) continuous on the entire real number line are a = 2 and b = 3.

`f(x)=\begin{cases}2x^2+x-3 \;\;&x\leq 2\\2x+3&2 < x < 5\\4x-7&x\geq 5\end{cases}`

Now, evaluate the limit as x approaches 3 using these values.

Since we have already ensured that the function is continuous, we can simply substitute x = 3 into the second case because 3 falls into the interval 2 < x < 5:

`\displaystyle\lim_(x \to 3)f(x)= \lim_(x \to 3) (2x+3)=2(3)+3=9`

Therefore, the limit of f(x) as x approaches 3 is 9.