Question

How do you find the constant a so that the function is continuous on the entire real line given `f(x)=4,`x <= -1`, ax +b, -1< x <1 and 6,`x>=1#?

Answer 1

We have `a=1` and `b=5` giving:

`f(x)={ (4,x<=-1), (x+5,-1 <= x <= 1), (6,x>=1) :}`

Explanation:

We want to find `a` and `b` such that `f(x)` is continuous:

`f(x)={ (4,x<=-1), (ax+b,-1 < x < 1), (6,x>=1) :}`

Strictly speaking we want to find

`f(x)={ (4,x<=-1), (ax+b,-1 <= x <= 1), (6,x>=1) :}`

Just think about what we know so far, and how it would look:

So for the mid interval we need a straight line passing through `(-1,4)` and `(1,6)`

This line would have the following gradient:
`m=(Delta y)/(Delta x) = (6-4)/(1-(-1)) = 2/2=1`
Hopefully you could also establish that by inspection!

So our required line passes through `(1,6)` (equally we could you the other coordinate and get the same answer) and has gradient `m=1`, so using `y-y_1=m(x-x_1)` the equation is:

`y -6 = (1)(x - 1)`
`:. y - 6 = x - 1`
`:. y = x+5`

Which we can graph to confirm

Hence, we have `a=1` and `b=5` giving:

`f(x)={ (4,x<=-1), (x+5,-1 <= x <= 1), (6,x>=1) :}`