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For what value of the constant c is the function f continuous on (-∞, ∞)?

– Given Function

\[ \ f\left( x\right)= \bigg\{\begin{array}{rcl} cx^2+2x, & x<2 \\ x^3-cx, & x≥2 \end{array}\]

The aim of the question is to find the value of constant c for which the given function will be continuous on the whole real number line.

The basic concept behind this question is the concept of Continuous Function.

A function f is a continuous function at x=a if it full fills the following conditions:

\[f\left(a\right)\ exists\]

\[\lim_{x\rightarrow a}{f(x)\ exists}\]

\[\lim_{x\rightarrow a}{f(x)\ =\ f(a)}\]

If the function is continuous at all the given points in an interval $(a,\ b)$, it is classified as a Continuous Function on the interval $(a,\ b)$

Expert Answer

Given that:

\[ \ f\left( x\right)= \bigg\{\begin{array}{rcl} cx^2+2x, & x<2 \\ x^3-cx, & x≥2 \end{array}\]

We know that if $f$ is a continuous function, then it will be also continuous at $x=2$.

\[ \lim_ { x \rightarrow 2^{+}}\ \ {f\left(x\right)\ }=\ \lim_{x\rightarrow2^-}\ \ {f\left(x\right)\ }=\lim_{x\rightarrow2}\ \ {f\left(x\right)\ }=\ {f\left(2\right)\ } \]

\[ \lim_{x\rightarrow2^-}\ \ {f\left(x\right)\ }=\ cx^2+2x \]

We know that $x<2$ so, to see if the function is continuous at $x=2$ put the value of $x$ here equal to $2$.

\[ \lim_{x\rightarrow2^-}\ \ {f\left(x\right)\ }=\ c{(2)}^2+2(2) \]

\[ \lim_{x\rightarrow2^-}\ \ {f\left(x\right)\ }=\ 4c+4 \]

Now, for the other equation, we have:

\[ \lim_{x\rightarrow2^+}\ \ {f\left(x\right)\ }=\ x^3-cx \]

We know that $x\le2$ so putting to see if the function is continuous at $x=2$ put the value of $x$ here equal to $2$.

\[ \lim_{x\rightarrow2^+}\ \ {f\left(x\right)\ }=\ {(2)}^3-c(2) \]

\[ \lim_{x\rightarrow2^+}\ \ {f\left(x\right)\ }=\ 8-2c \]

From the above equations, we know that:

\[ \lim_{x\rightarrow2^-}\ \ {f\left(x\right)\ }=\ \lim_{x\rightarrow2^+}\ \ {f\left(x\right)\ } \]

Putting values of both limits here, we get:

\[ 4c+4 = 8-2c \]

\[ 4c-2c = 8-4 \]

\[ 6c = 4 \]

\[ c =\frac{4}{6} \]

\[ c =\frac{2}{3} \]

From above equation we find out the value of Constant $c$ for the given Continuous Function:

\[ c =\frac{2}{3} \]

Numerical Result

So the value of constant $c$ for which the given function $ \ f\left( x\right)= \bigg\{\begin{array}{rcl} cx^2+2x, & x<2 \\ x^3-cx, & x≥2 \end{array}$  is continuous on the whole real number line is as follows:

\[ c =\frac{2}{3} \]

Example

Find out the value of constant $a$ for the given continuous function:

\[\ f\left( x\right)= \bigg\{ \begin{array}{rcl} x3, & x≤4 \\ ax^2, & x>4 \end{array}\]

Solution

We know that if $f$ is a continuous function, then it will also be  continuous at $x=4$.

\[ \lim_ { x \rightarrow 4^{+}}\ \ {f\left(x\right)\ }=\ \lim_{x\rightarrow4^-}\ \ {f\left(x\right)\ }=\lim_{x\rightarrow4}\ \ {f\left(x\right)\ }=\ {f\left(4\right)\ }\]

\[ \lim_{x\rightarrow4^+}\ \ {f\left(x\right)\ }=\ ax^2 \]

\[ \lim_{x\rightarrow4^+}\ \ {f\left(x\right)\ }=\ a{(4)}^2 \]

\[ \lim_{x\rightarrow4^+}\ \ {f\left(x\right)\ }=\ 16a \]

\[ \lim_{x\rightarrow4^-}\ \ {f\left(x\right)\ }=\ x^3 \]

\[ \lim_{x\rightarrow4^-}\ \ {f\left(x\right)\ }=\ {(4)}^3 \]

\[ \lim_{x\rightarrow4^-}\ \ {f\left(x\right)\ }=\ 64 \]

From the above equations, we know that:

\[ \lim_{x\rightarrow4^-}\ \ {f\left(x\right)\ }=\ \lim_{x\rightarrow4^+}\ \ {f\left(x\right)\ } \]

Equating both equations:

\[16a=64\]

\[a=\frac {64}{16}\]

\[a=4\]

Hence, the value of Constant $a$ is:

\[a=4\]

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