Answer:
-3/2
The problem:
Find [tex]c[/tex] so the function [tex]f[/tex] is continuous.
[tex]f(x)=cx^2+6x, x<2[/tex]
[tex]f(x)=x^3-x,x \ge 2[/tex]
Please read my interpretation of the problem above.
Step-by-step explanation:
If we want [tex]f[/tex] continuous at [tex]x=a[/tex], then we want the following things:
[tex]\lim_{x \rightarrow a}f(x)=f(a)[/tex] (This means the left and right limits at [tex]x=a[/tex] need to equal.)
Of course the limit and [tex]f(a)[/tex] need to both be a number.
So we want to basically solve the following equation:
[tex]c(2)^2+6(2)=(2)^3-(2)[/tex]
[tex]c(4)+12=8-2[/tex]
[tex]4c+12=6[/tex]
Subtract 12 on both sides:
[tex]4c=-6[/tex]
Divide both sides by 4:
[tex]c=\frac{-6}{4}[/tex]
Reduce:
[tex]c=\frac{-3}{2}[/tex]
