Question

A real-valued function:$f(x) = C{\log _e}|x| + D{x^3} + x,x \ne 0$, where C and D are constants, has critical points at $x = - 1$ and$x = 2$. Then the ordered pair of $\left( {C,D} \right)$ is

Answer 1

Hint:First find the derivative of the function and then use the definition of the critical points to find the critical points of the given function at $x = - 1$ and $x = 2$ and finally find the ordered pairs.

Complete step-by-step answer:
We have given a function $f(x) = C{\log _e}|x| + D{x^3} + x,x \ne 0$, where C and D are constants and the function has the critical point at $x = - 1$ and$x = 2$.
We have to find the value of the ordered pair $\left( {C,D} \right)$.
$f(x) = C{\log _e}|x| + D{x^3} + x,x \ne 0$
First, we find the differentiation of the given function:
We know differentiation of log x is $\dfrac{1}{x}$ and $x^n= nx^{n-1}$
Using these formulas we try to differentiate the function,
$f'(x) = \dfrac{C}{x} + 3{x^2}D + 1$
We know that the critical point of the function is obtained by setting the derivative of the function equals to zero. That is,
$f'\left( x \right) = 0$
Substitute the value of the derivative of the function:
$\dfrac{C}{x} + 3{x^2}D + 1 = 0$
It is given that the function has a critical point at $x = - 1$, so put the value of $x$ into the above equation:
$ - \dfrac{C}{1} + 3{\left( 1 \right)^2}D + 1 = 0$
$ \Rightarrow - C + 3D + 1 = 0$ … (1)
It is given that the function has a critical point at $x = 2$, so put the value of $x$into the above equation:
$ - \dfrac{C}{2} + 3{\left( 2 \right)^2}D + 1 = 0$
$ \Rightarrow \dfrac{C}{2} + 12D + 1 = 0$
Multiply with 2 on both sides:
$C + 24D + 2 = 0$ … (2)
Add equations 1 and 2 we get:
$ - C + 3D + 1 + C + 24D + 2 = 0$
$ \Rightarrow 27D + 3 = 0$
$ \Rightarrow D = \dfrac{{ - 3}}{{27}}$
$ \Rightarrow D = - \dfrac{1}{9}$
Now, we put the value of D into the equation (1):
$ - C + 3\left( { - \dfrac{1}{9}} \right) + 1 = 0$
$ \Rightarrow - C - \dfrac{1}{3} + 1 = 0$
$ \Rightarrow C = 1 - \dfrac{1}{3} = \dfrac{2}{3}$
Now we have the values $D = - \dfrac{1}{9}$ and $C = \dfrac{2}{3}$.
Therefore, the ordered pair $\left( {C,D} \right)$ has the value $\left( { - \dfrac{1}{9},\dfrac{2}{3}} \right)$.


Note:Critical points are the points at which the derivative of the function is zero. The critical point of the function is applicable in finding the relative maxima, relative minima, and the concavity of the function. It is also used in finding the interval in which the function is increasing and decreasing.