( f(x)=lefteginarrayllc sin left(fracpi x2 ight) x leq...

Transcript

00:01 In this problem we need to determine the value of the constant c for which the given function f is continuous from negative infinity to infinity.

00:08 So the value of c for which the function is continuous for all real numbers.

00:13 Now note that f of x is equal to cx squared plus 2x when x is less than 3.

00:19 And this is a polynomial function.

00:23 Now when x is greater than or equal to 3, f of x is x cubed minus cx.

00:30 This is also a polynomial function.

00:33 And polynomial functions are continuous for all real numbers.

00:36 So no matter what the value of c may be, it will be this function fx will be continuous for all values of x which is not equal to 3.

00:46 Why are we excluding 3? because 3 is the boundary point.

00:49 We have x is less than 3 and x is greater than or equal to 3.

00:52 So excluding this boundary point, we can say that it is continuous for all real numbers.

00:57 So we want it to be continuous when x is 3 as well.

01:01 So for that, we will have the limit as x tends to 3 minus f of x.

01:06 This needs to be equal to the limit as x tends to 3 plus f of x.

01:10 And this needs to be equal to the value of f at 3.

01:14 So the limit as x tends to 3 minus f of x.

01:17 3 minus, that means x approaches 3 from the left.

01:21 X is less than 3 and when x is less than 3, f of x is cx squared plus 2x.

01:27 Then we have the limit as x tends to 3 plus f of x...

Question

( f(x)=left{egin{array}{ll}c sin left(frac{pi x}{2} ight) & x leq 3 \ x^{2}+c x-1 & x>3 .end{array} ight. )

Question

( f(x)=left{egin{array}{ll}c sin left(frac{pi x}{2} ight) & x leq 3 \ x^{2}+c x-1 & x>3 .end{array} ight. )

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( f(x)=left{egin{array}{ll}c sin left(frac{pi x}{2} ight) & x leq 3 \ x^{2}+c x-1 & x>3 .end{array} ight. )