Continuity of a piece wise function

Question

Let a and b represent real numbers. Define

f(x)=

ax^2+x-b if x <= 2

ax+b if 2<x<5

2ax-7 if x >= 5

1) How to find values of a and b such that f is continuous everywhere?

Answer 1

In this example, we have 3 different polynomial functions defined on 3 different intervals.

Any polynomial function is continuous, so we don't have to worry about any discontinuities within the intervals that those functions are defined.

Therefore, we can say with certainty that the function is continuous on the intervals (-∞, 2), (2, 5), and (5, ∞). Note that I used parentheses rather than brackets, so that means the intervals of continuity we are certain of do not include the endpoints. ∞ and -∞ are covered because they don't end. So, in order for the whole function to be continuous, we need to make sure that our graph connects at those points where x = 2 and x = 5.

How can we determine the continuity at those points? Well at x =2, the point that the function approaches from the left is defined by f(x)=ax^2+x-b, and from the right, it's f(x)= ax + b.

so, we need to find the y value that these functions would reach as they got to x = 2. We can do that by plugging in 2.

from the left f(x) = 4a + 2 - b

from the right f(x) = 2a + b

In order for the function to be continuous, these y values need to be equal.

Similarly, we need the other end point (x = 5) to be the same way.

so f(x) = ax+b is on the left for this end point, and f(x) = 2ax-7 is on the right.

then we plug in 5.

we get 5a + b = 10a - 7 if we set them equal.

We now have a system of 2 equations with two unknowns. The solution to this system will give us the answer we are looking for.