find the constants a and b so that the function is continuous on the entire real line. F(x)= 3, x≤-1
ax+b, -1<x<2
-3,x≥2 Help???
A little more legibly,
f(x)
= 3 for x<=1
= ax+b for 1<x<2
= -3 for x>=2
So, you need to find a line connecting (-1,3) with (2,-3) so there are no holes in the graph.
That's just a simple two-point line problem:
(y-3)/(x+1) = -6/3 = -2
y-3 = -2(x+1)
y = -2x + 1
So, f(x) = -2x+1 for -1<x<2
To find the constants a and b such that the function is continuous on the entire real line, we need to ensure that the function is continuous at the points where the function changes its definition.
In this case, the function changes at x = -1 and x = 2.
At x = -1, the function is defined as ax + b for -1 < x < 2. We need to ensure that the left and right limits of the function are equal at this point.
Taking the left limit as x approaches -1: lim(x→-1-) (ax + b) = a(-1) + b = -a + b
Taking the right limit as x approaches -1: lim(x→-1+) (ax + b) = a(-1) + b = -a + b
For the function to be continuous at x = -1, these two limits need to be equal. Therefore, -a + b = -a + b.
Moving on to x = 2, the function is defined as -3 for x ≥ 2. Again, we need to ensure that the left and right limits of the function are equal at this point.
Taking the left limit as x approaches 2: lim(x→2-) (-3) = -3
Taking the right limit as x approaches 2: lim(x→2+) (-3) = -3
Since both limits are equal, the function is already continuous at x = 2.
Therefore, the only condition we need to satisfy is -a + b = -a + b. This equation is always true regardless of the values of a and b. So any values of a and b will make the function continuous on the entire real line.
In summary, the function F(x) = 3 for x ≤ -1, ax + b for -1 < x < 2, and -3 for x ≥ 2 is continuous on the entire real line for any values of a and b.