Question

How do you find the stationary and inflection points for ${{x}^{2}}-2x-3$?

Answer 1

To find the stationary point of the given function, we have to equate y with the function, that is $y={{x}^{2}}-2x-3$. Then, we have to differentiate the whole equation with respect to y. Assume the value of $\dfrac{dy}{dx}$ to be zero. Do necessary calculations and find the value of x. Once we find the value of x, we have to substitute the value of x in the equation $y={{x}^{2}}-2x-3$ and find the value of y. thus, the stationary point is obtained. To find the inflection points, we have to double differentiate y. We get 2, which is a constant and never changes signs. Thus, there are no inflection points for the given function.

Complete step by step solution:
According to the question, we are asked to find the stationary point and inflection points of the function ${{x}^{2}}-2x-3$.
We have been given the function is ${{x}^{2}}-2x-3$.
Let us assume $y={{x}^{2}}-2x-3$. ------------(1)
We first have to find the stationary points.
Let us differentiate the function (1) with respect to x.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{2}}-2x-3 \right)$
Using the addition rule of differentiation $\dfrac{d}{dx}\left( u+v+w \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}+\dfrac{dw}{dx}$, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( -2x \right)+\dfrac{d}{dx}\left( -3 \right)$
We know that the differentiation of a constant is 0.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( -2x \right)$
We can further simplify the differentiation as
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{2}} \right)-2\dfrac{dx}{dx}$
We know that $\dfrac{dx}{dx}=1$. Using this in the above differentiation, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{2}} \right)-2\left( 1 \right)$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{2}} \right)-2$
Using the power rule of differentiation, that is $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$, we get
$\dfrac{dy}{dx}=2{{x}^{2-1}}-2$
$\Rightarrow \dfrac{dy}{dx}=2x-2$ -----------(2)
Let us now substitute $\dfrac{dy}{dx}=0$. We get
2x-2=0
Let us now add 2 on both the sides.
$\Rightarrow 2x-2+2=0+2$
On further simplification, we get
$2x=2$
On dividing the whole expression by 2, we get
$\Rightarrow \dfrac{2x}{2}=\dfrac{2}{2}$
We find that 2 are common in the numerator and denominator of the LHS and RHS. On cancelling 2, we get
x=1
Now, we have to find the value of y by substituting the value of x in equation (1).
$\Rightarrow y={{\left( 1 \right)}^{2}}-2\left( 1 \right)-3$
On further simplifications, we get
$y=1-2-3$
$\Rightarrow y=1-5$
$\therefore y=-4$
Therefore, we get the stationary point which is (1,-4).
Now, we have to find the inflection points.
To find the inflection points, we have to double differentiate equation (1) or differentiate equation (2) with respect to x.
Let us differentiate (2) with respect to x.
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 2x-2 \right)$
Using the addition rule of differentiation, that is $\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}$, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 2x \right)+\dfrac{d}{dx}\left( -2 \right)$
We know that differentiation of a constant is 0.
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 2x \right)$
We can express the differentiation as
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2\dfrac{dx}{dx}$
We know that $\dfrac{dx}{dx}=1$. Therefore, we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2$
We find that the double differentiation is a constant and does not change the sign on changing the value of x.
Therefore, we get that there is no inflection point in the function.
Hence, the stationary point of ${{x}^{2}}-2x-3$ is (1,-4) and there are no inflection points.

Note: Whenever we get such types of problems, we have to equate the given function to y or f(x) and then solve. Do the differentiations carefully. Avoid calculation mistakes based on sign conventions. We should always equate the differentiation of y to 0 to find the stationary points.