For what value of the constant c is the function f(x) continuous for all real numbers? f(x) = \begin{cases} cx^2 + 2x \text{ if } x \ \textless \ 2\\ x^3 -cx \ \, \text{ if } x …

Question

For what value of the constant c is the function f(x) continuous for all real numbers?


`f(x) = \begin{cases} cx^2 + 2x \text{ if } x \ \textless \ 2\\ x^3 -cx \ \, \text{ if } x \ge 2\end{cases}`

Answer 1

Explanation:

For the function to be continuous for all real numbers, it must be continuous at x = 2.

To find the value of c that makes the function continuous at x = 2, we need to ensure that the left-hand limit and the right-hand limit of the function are equal at x = 2, and that they are also equal to the function value at x = 2.

The left-hand limit is:

lim x → 2- f(x) = lim x → 2- (cx^2 + 2x) = c(2^2) + 2(2) = 4c + 4

The right-hand limit is:

lim x → 2+ f(x) = lim x → 2+ (x^3 - cx) = 2^3 - c(2) = 8 - 2c

The function value at x = 2 is:

f(2) = c(2^2) + 2(2) = 4c + 4

For the function to be continuous at x = 2, the left-hand limit, right-hand limit, and function value must be equal. Therefore, we can set up the following equation:

4c + 4 = 8 - 2c

Solving for c, we get:

6c = 4

c = 4/6

c = 2/3

Therefore, the value of the constant c that makes the function continuous for all real numbers is c = 2/3.