Question

For what value of the constant c is the function fcontinuous on (−[infinity], [infinity])?

f(x) = cx^2 + 5x if x<6
x^3 - cx if x>_6
Note that f is continuous on (−[infinity], 6) and (6, [infinity]). For the function to be continuous on (−[infinity], [infinity]), we need to ensure that as x approaches 6,the left and right limits match.

Answer 1

For
`c=(1)/(7)` the function f(x) is continuous on
`(-\infty,\infty)`.

Explanation:

We have the following function

`f(x) = \left\{ \begin{array}{ll} cx^2+5x & \quad x <6 \\ x^3-cx & \quad x \geq 6 \end{array} \right.`

For the function f(x) to be continuous on
`(-\infty,\infty)` it is sufficient to have continuity at x = 6, we need to ensure that as x approaches 6, the left and right limits match, this means that

`\lim_(x \to 6^(-) ) f(x)=\lim_(x \to 6^(+) ) f(x)=f(x)`,

which holds if and only if

`c\left(6\right)^2+5\left(6\right)=\left(6\right)^2-c\left(6\right)\\36c+30=36-6c\\42c=6`

namely if
`c=(1)/(7)`.