– 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:
\[a=4\]