For what values of a and b that make the function f continuous everywhere.
f(x)= {x−2x2−4 if x<2
ax2−bx−18 if 2< or =x<3
10x−a+b if x> or =3
look at 2
x^2-2x-4=ax^2-bx-18
4-4-4=4a-2b-18
2a-b=-18 check that.
at x=3
ax^2-bx-18=10x-a+b
9a-3b-18=30-a+b
10a-4b=48
Now you have two equations, two unknowns, solve.
To find the values of 'a' and 'b' that make the function 'f' continuous everywhere, we need to ensure that the function is continuous at each transition point, which occurs when the conditions change for the different intervals.
First, we consider the transition from the first interval to the second interval at x = 2. For the function to be continuous at x = 2, the left-hand limit as x approaches 2 (from the first interval) must be equal to the right-hand limit as x approaches 2 (from the second interval). Therefore, we need to find the left-hand limit and the right-hand limit at x = 2.
Left-hand limit as x → 2:
lim(x → 2-) f(x) = lim(x → 2-) (x - 2x^2 - 4)
To calculate this, we substitute x = 2 into the function:
lim(x → 2-) f(x) = 2 - 2(2^2) - 4 = 2 - 2(4) - 4 = -6
Right-hand limit as x → 2:
lim(x → 2+) f(x) = lim(x → 2+) (ax^2 - bx - 18)
To calculate this, we substitute x = 2 into the function:
lim(x → 2+) f(x) = a(2^2) - b(2) - 18 = 4a - 2b - 18
For the function to be continuous at x = 2, these two limits must be equal:
-6 = 4a - 2b - 18
Simplifying this equation, we have:
4a - 2b = 12 -- (Equation 1)
Now, let's consider the transition from the second interval to the third interval at x = 3. Again, we need to find the left-hand limit and the right-hand limit at x = 3.
Left-hand limit as x → 3:
lim(x → 3-) f(x) = lim(x → 3-) (ax^2 - bx - 18)
To calculate this, we substitute x = 3 into the function:
lim(x → 3-) f(x) = a(3^2) - b(3) - 18 = 9a - 3b - 18
Right-hand limit as x → 3:
lim(x → 3+) f(x) = lim(x → 3+) (10x - a + b)
To calculate this, we substitute x = 3 into the function:
lim(x → 3+) f(x) = 10(3) - a + b = 30 - a + b
For the function to be continuous at x = 3, these two limits must be equal:
9a - 3b - 18 = 30 - a + b
Simplifying this equation, we have:
10a - 4b = 48 -- (Equation 2)
Now, we have a system of two equations (Equation 1 and Equation 2) with two unknowns (a and b). We can solve this system to find the values of 'a' and 'b' that make the function 'f' continuous everywhere.
By solving Equation 1 and Equation 2 simultaneously, the values of 'a' and 'b' can be determined.