Find the Values for which the function is continuous

[mazz1801]

Homework Statement

Determine the values of k,L,m and n such that the following function g(x) is continuous and differentiable at all points

Homework Equations

2x²-n if x<-2
mx+L if -2≤x<2
kx²+1 if x≥2

The Attempt at a Solution

So I know that for the function to be continuous the limit as x→c of f(x) must equal f(c)
And I am trying to make this function continuous for all values of k,L,m and n.

so I done the following
2x²-n = mx+L for x=-2
mx+L = kx²+1 for x= 2

and I simplified them to the following.
2m-n+5=0
2m-4k+L=0


From here I don't know what to do... I'm pretty sure that I am correct up to here but could someone just please tell me what to do now...
I don't know how to solve for 4 variables with only 2 equations

 

[andrien]

just differentiate and then equate them to find.four eqn four unknown

 

[HallsofIvy]

Homework Statement

Determine the values of k,L,m and n such that the following function g(x) is continuous and differentiable at all points

Homework Equations

2x²-n if x<-2
mx+L if -2≤x<2
kx²+1 if x≥2

You should know that all polynomials are both continuous and differentiable for all x so that only points in question are x= -2 and x= 2

The Attempt at a Solution

So I know that for the function to be continuous the limit as x→c of f(x) must equal f(c)
And I am trying to make this function continuous for all values of k,L,m and n.

so I done the following
2x²-n = mx+L for x=-2
mx+L = kx²+1 for x= 2

and I simplified them to the following.
2m-n+5=0

What happened to the "L"? At x= -2, the first equation becomes
2(-2)²- n= m(-2)+ L which simplifies to 8- n= -2m+ L or 2m- n+ 8- L= 0.

2m-4k+L=0

And here, what happened to the "1"? At x= 2, the second equation becomes m(2)+ L= k(4)+ 1 or 2m- 4+ L- 1= 0

From here I don't know what to do... I'm pretty sure that I am correct up to here but could someone just please tell me what to do now...
I don't know how to solve for 4 variables with only 2 equations

You have not yet said any thing about being "differentiable".

(Note, the derivative of a function is NOT necessarily differentiable but it does satisfy the "intermediate value property" so IF the limit from each side exist, then they must be equal.)