How to find the value of a missing variable to make a function continuous

Question

Determine the value of A that will make the function continuous at x=3

 

f(x)={ 3x^2-5x+A     If A >=1

           5x+2A            If A <1

Answer 1

3*3^2-5*3+A = 5*3+2A

 

27 - 15 + A = 15 + 2A

 

A = -3

 

But then f(x) = 5x-6 b/c A <1;

 

and f(x) = 5x-6 is continuous everywhere.

Answer 2

Can you double check the question? f(x) is continuous at x=3 for all values of A.

Answer 3

Hi Paige;

f(x)=3x²-5x+A, if A>=1

5x+2A, if A<1

A cannot be both greater than and less than 1.

Otherwise, this is my solution...

3x²-5x+A=5x+2A

Let's bring everything together by subtracting (5x+2A) from both sides...

3x²-10x-A=0

We must factor.

For the FOIL, one parenthetical equation must be (x-3) for the function to be continuous at x=3...

FIRST must be (3x)(x)=3x²

OUTER must be (3x)(-3)=-9x

INNER must supplement OUTER such that the total is -10x...-9x+-1x=-10x

INNER must be (-1)(x)=-1x=-x

LAST must be (-1)(-3)=3

(3x-1)(x-3)=0

Let's FOIL...

FIRST...(3x)(x)=3x²

OUTER...(3x)(-3)=-9x

INNER...(-1)(x)=-1x=-x

LAST...(-1)(-3)=3

3x²-9x-x+3=0

3x²-10x+3=0

The equation was...

3x²-10x-A=0

3x²-10x--3=0

A=-3

 

Let's check our work...

3x²-5x+A=5x+2A

A=-3

3x²-5x-3=5x+[(2)(-3)]

3x²-5x-3=5x-6

x=3

[(3)(3²)]-[(5)(3)]-3=[(5)(3)]-6

[(3)(9)]-[(5)(3)]-3=[(5)(3)]-6

27-15-3=15-6

27-18=15-6

9=9