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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 :

Answer:

For [tex]c=\frac{1}{7}[/tex] the function f(x) is continuous on [tex](-\infty,\infty)[/tex].

Step-by-step explanation:

We have the following function

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

For the function f(x) to be continuous on [tex](-\infty,\infty)[/tex] 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

[tex]\lim_{x \to 6^{-} } f(x)=\lim_{x \to 6^{+} } f(x)=f(x)[/tex],

which holds if and only if

[tex]c\left(6\right)^2+5\left(6\right)=\left(6\right)^2-c\left(6\right)\\36c+30=36-6c\\42c=6[/tex]

namely if [tex]c=\frac{1}{7}[/tex].

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