Find the values of c and d that make the function continuous.

Question

F(x) = 4x {x<1}

F(x) = cx^2+d {1≤x<2}

F(x) = 7x {x≥2}

 

If anyone can help me out with this that would be amazing!

Answer 1

A continuous function is a function that has no gaps. You could draw it without picking up your pencil. The pertinent points are at the boundaries with x = 1 and x = 2.

 

First, x = 1

 

f(x) = 4x

f(1) = 4

 

f(x) = cx² + d also has to equal 4 at x = 1

4 = c(1)²+d

4 = c + d

 

Now let's look at x = 2

f(x) = 7x

f(2) = 7×2

f(2) = 14

 

f(x) = cx2 + d also has to equal 14 at x = 2

14 = c(2)² + d

14 = 4c + d

 

Now we have a system of equations with two variables.

4 = c + d

14 = 4c + d

 

There are many methods to solve this. I am going to use substitution.

 

4 = c + d                          Subtract c from both sides

4 - c = d

 

d= 4 - c

14 = 4c + d                      Substitute for d

14 = 4c + 4 - c                  Combine like terms

14 = 3c + 4                      Subtract 4 from both sides

10 = 3c                             Divide both sides by 3

10/3 = c

 

Go back to the original equations and substitute 10/3 for c

 

4 = c + d

4 = 10/3 + d                             Change 4 to thirds

12/3 = 10/3 + d                        Subtract 10/3 from both sides

2/3 = d

 

f(x) = 10/3 x² + 2/3

Answer 2

 

 Well the quadratic function cx^2 + d must have (1,4) and (2,14) because of the 4x and 7x

 

   So then    c+d = 4    and 4c +d = 14

 

                  Subtracting the first equation FROM the second gives  3c = 10

                  so c=10/3

 

 

    10/3 + d = 4 = 12/3

   d = 2/3

 

So the function is (10/3)x^2 + 2/3

 

 

BUT....

the derivatives must also match !!

 

SO... at the first mesh point, we have 4 = 2cx with x=1

                                                       4 = 2c

                                                       c =2

 

at the second mesh point we have 7 = 2cx with x=2

                                                  7 = 2c(2)

                                                 7 = 4c

                                                    c = 7/4

 

THe problem is the constant D disappears and remains undetermined

and these values contradict the ones found for the function value