Answer:
1. not continuous, as the function definitions deliver different function values at x=1 when approaching this x from the left and from the right side.
2.
2 = a + b
3.
0 = 2a + b
4.
a = -2
b = 4
Step-by-step explanation:
the function is continuous at a specific point or value of x, if the f(x) = y functional value is the same coming from the left and the right side at that point.
1. that means that for x=1
3 - x = ax² + bx
so,
3 - 1 = a×1² + b×1 = a + b
2 = a + b
we have to use a=2 and b=3
2 = 2 + 3 = 5
2 is not equal 5, so the assumed equality is false, so the function is not continuous there.
2. point 1 gave us already the working relationship between a and b.
2 = a + b
only if that is true, is the function continuous at x=1.
3. now for x=2
5x - 10 = ax² + bx
5×2 - 10 = a×2² + b×2 = 4a + 2b
10 - 10 = 4a + 2b
0 = 4a + 2b
0 = 2a + b
4. to find a and b to be continuous at both locations x=1 and x=2 both expressions in a and b must apply.
so, they establish a system of 2 equations with 2 variables.
2 = a + b
0 = 2a + b
a = 2 - b
0 = 2×(2-b) + b = 4 - 2b + b = 4 - b
b = 4
therefore
a = 2 - 4 = -2
5. I cannot draw a graph here.
just use now the function
3 - x, x < 1
‐2x² +4x, 1 <= x < 2
5x - 10, x >= 2
