Continuous functions Beginning cal?

Question

For what value of the constant c is the function f continuous on (-infinant , infinant )?
f(x)={cx^2-3x if x < 1
___{x^3-cx if x >=1

I dont really understand how to do this can someone please explain it to me.

the problem looks like
{ cx^2 + 3x if x<1
f(x)= {
{ x^3 - cx if x>=1

2 does not seem to be the answer I tried

Sorry guys both of you are wrong the answer is -1

Cx^2+3x=X^3-CX
Put 1 inplace of x
C+3=1-C
C+3-1+C=0
2C+2=0
2C=-2
C=-2/2
C=-1

I did some reasearch and found out that you set each one = to the other replace x With 1 and solve for c

Answer 1

f(x)={cx^2-3x if x < 1

f(x) = x^3-cx if x >=1

f(1)=1-c

f(1+)= 1-c
f(1-)= c-3
1-c=c-3
1+3=2c
4=2c
c=4/2=2

Answer 2

Whoops, misunderstood the problem.

We must make sure that at x =1 the function is continuous.
This requires that the two functions have the same slope at at x=1. The slope of the 1st part of the function is 2xc-3 = 2c-3 when x =1. The slope of the second part of the function is 3x^2-c = 3-c at x=1
So 3-c = 2c-3
3c = 6
c=2