## Find the Values for which the function is continuous

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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
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 just differentiate and then equate them to find.four eqn four unknown

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 Quote by mazz1801 1. The problem statement, all variables and given/known data Determine the values of k,L,m and n such that the following function g(x) is continuous and differentiable at all points 2. Relevant 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

 3. 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.)

 Tags continuous function, unknown, variables

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