Find all values of a and b that make f(x) continuous everywhere.

Question

f(x) = {ax^2 +bx+ 4 when x < 1

{ bx^3 -2 when 1 < x < 3

{ 25 when x > 3

Answer 1

I will start by solving for b first:

The third part of your function says that f(x) = 25 for x > 3 and f(x) = bx³ - 2 for x < 3. In order for it to be continuous, both parts have to be equal at x = 3:

25 = bx³ - 2

Plug in x = 3 and solve for b:

25 = b(3)³ - 2

25 =27b - 2

27 = 27b

b = 1

So, now we can plug in b into all three parts of the function:

f(x) = ax² + x + 4, x < 1

= x³ - 2, 1 < x < 3

= 25, x ≥ 3

In order to find the value of a, the first two parts of f(x) have to be equal at x = 1:

ax² + x + 4 = x³ - 2

plug in x = 1

a + 1 - 4 = 1 - 2

a - 3 = - 1

a = 2

So, our values are:

a = 2

b = 1

Answer 2

The function is broken into 3 pieces like this:

I show the points that separate the pieces as A and B. If the function is to be continuous then piece #1 must end at the same place piece #2 starts (which is at point A. We know that the x-value of point A is 1. So, we know that the y-value of piece #1, when x = 1, must be the same as the y-value of piece #2, when x = 1. So a(1)² +b(1)+ 4 must equal b(1)³ - 2 or

a + b + 4 = b - 2 or

a + 4 = -2

a = -6

The same holds true for point B.

b(3)³ - 2 = 25

27b - 2 = 25

27b = 27

b = 1

Answer 3

Set ax^2 + bx + 4 = bx^3 -2 when x = 1 and bx^3 -2 = 25 when x = 3

We want to connect these the pieces of the function so that they all bump right into each other at the respective endpoints (=) or limit points (< or >), then the function won’t have any gaps and will be continuous.

so b = 1 from second equation above, and then a + 1 + 4 = -1 from first equation, so a = -6