Question

Do not use your calculator for Parts 1 through 4.

Let f be the function defined as follows:

{ 3-x, for x < 1
f(x) = { ax^2 + bx, for 1 ≤ x < 2
{ 5x-10, for x ≥ 2

where a and b are constants

The equation in written form: f of x equals the piecewise function three minus x for x is less than one, a x squared plus b x for one is less than or equal to x is less than two, and five x minus ten for x is greater than or equal to two where a and b are constants.

1. If a = 2 and b = 3, is f continuous at x = 1? Justify your answer.

2. Find a relationship between a and b for which f is continuous at x = 1.
Hint: A relationship between a and b just means an equation in a and b.

3. Find a relationship between a and b so that f is continuous at x = 2.

4. Use your equations from parts (ii) and (iii) to find the values of a and b so that f is continuous at both x = 1 and also at x = 2?

5. Graph the piece function using the values of a and b that you have found. You may graph by hand or use your calculator to graph and copy and paste into the document.

Answer 1

you have to understand how piecewise functions first
I hope you do because I won't be explaining it here



1.
if it is continiuous at x=1 then the part of the function that approaches 1 from the left is the same value
so we have f(x)=3-x for x<1, so evaluate it for x=1,
we would get 3-1=2
so
if a=2 and b=3

f(x)=2x²+3x
for x=1, we get 2+3=5

it approaches 2 from the left but equals 5
2≠5
so it is not continuous because it doesn't approach the same value at x=1




2.
alright
so we established that it should approach 2 because we had the f(x)=3-x
so
f(1)=2=ax²+bx
2=a+b
the relationship is 2=a+b



3.
same as before, but use the other function, the one that is defined at x≥2
5x-10, 5(2)-10, 10-10, 0
it approaches 0
so
we must find one such that f(x)=0 for x=2 with the ax²+bx
0=a(2)²+b(2)
0=4a+2b
if we minus 2b both sides
-2b=4a
divide by 2
-b=2a
add b both sides
0=2a+b



4.
2=a+b
0=2a+b

2=a+b
minus b both sides
2-b=a
subsitute
0=2a+b
0=2(2-b)+b
0=4-2b+b
0=4-b
b=4

sub back
2-b=a
2-4=a
-2=a

a=-2
b=4


5.
graph f(x)=-2x²+4x from x=1 to x=2 and put a closed dot at (1,2) and an open dot at (2,0)

Answer 2

(i) Not continuous.

(ii) a + b = 2

(iii) 2a + b = 0

(iv) a = -2 and b = 4

(v) Graph attached.

Explanation:

Given piecewise function:

`f(x)=\begin{cases}3-x\;\;&\text{for}\;x < 1\\ax^2+bx\;\;&\text{for}\;1 \leq x < 2\\5x-10\;\;&\text{for}\;x\geq2\end{cases}`

where
`a` and
`b` are constant.

`\hrulefill`

Question (i)

If a = 2 and b = 3, then:

`f(x)=2x^2+3x,\;\;\text{for}\;1 \leq x < 2`

To determine if f is continuous at x = 1, we need to check if the limit of f(x) as x approaches 1 from both the left and the right is equal to f(1).

Value of f(1)

`f(1)=2(1)^2+3(1)=2+3=5`

Left-hand limit as x approaches 1:

`\displaystyle \lim_(x \to 1^(-)) f(x)=\displaystyle \lim_(x \to 1^(-)) (3-x)=3-1=2`

Right-hand limit as x approaches 1:

`\displaystyle \lim_(x \to 1^(+)) f(x)=\displaystyle \lim_(x \to 1^(+)) (2x^2+3x)=2+3=5`

Since the left-hand limit, right-hand limit, and the value of f at x = 1 are not equal, f is not continuous at x = 1 when a = 2 and b = 3.

`\hrulefill`

Question (ii)

To find a relationship between a and b for which f is continuous at x = 1, we need the left-hand limit, right-hand limit, and the value of f at x = 1 to be equal. Therefore, we set them equal to each other:

`\begin{aligned}3-x&=ax^2+bx\\3-1&=a(1)^2+b(1)\\2&=a+b\\a+b&=2\end{aligned}`

Therefore, the relationship between a and b for which f is continuous at x = 1 is a + b = 2.

`\hrulefill`

Question (iii)

To find a relationship between a and b so that f is continuous at x = 2, we need to check if the limit of f(x) as x approaches 2 from both the left and the right is equal to f(2).

Value of f(2)

`f(2)=5(2)-10=10-10=0`

Left-hand limit as x approaches 2:

`\displaystyle \lim_(x \to 2^(-)) f(x)=\displaystyle \lim_(x \to 2^(-)) (ax^2+bx)=a(2)^2+b(2)=4a+2b`

Right-hand limit as x approaches 2:

`\displaystyle \lim_(x \to 2^(+)) f(x)=\displaystyle \lim_(x \to 2^(+)) (5x-10)=5(2)-10=0`

To make f continuous at x = 2, the left-hand limit, right-hand limit, and the value of f at x = 2 should be equal. Therefore, we set them equal to each other:

`\begin{aligned}4a+2b&=0\\2(2a+b)&=0\\2a+b&=0\end{aligned}`

Therefore, the relationship between a and b for which f is continuous at x = 2 is 2a + b = 0.

`\hrulefill`

Question (iv)

To make f continuous at both x = 1 and x = 2, we need to satisfy both the relationships found in part (ii) and (iii).

The two equations are:

`a + b = 2`

`2a + b = 0`

To solve for a and b, begin by rearranging the first equation to isolate a:

`a=2-b`

Substitute this into the second equation and solve for b:

`2(2-b)+b=0`

`4-2b+b=0`

`4-b=0`

`b=4`

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

`a=2-b`

`a=2-4`

`a=-2`

Therefore, the values of a and b so that f is continuous at both x = 1 and x = 2 are a = -2 and b = 4.

`\hrulefill`

Question (v)

Now we have found the values of a and b, the piecewise function is:

`f(x)=\begin{cases}3-x\;\;&\text{for}\;x < 1\\-2x^2+4x\;\;&\text{for}\;1 \leq x < 2\\5x-10\;\;&\text{for}\;x\geq2\end{cases}`

Use a graphing calculator, we can graph the function (attached).

• The red line is the first piece of the function.
• The blue curve is the second piece of the function.
• The third line is the third piece of the function.