Question

Minimum value of the quadratic function ${x^2} - 3x + 5$ ?

Answer 1

Given problem tests the concepts of derivatives and their applications. In the problem, we are required to find the minimum value of the quadratic function in the variable x as ${x^2} - 3x + 5$. We do so by differentiating the function with respect to x and finding the critical points at which the function attains maximum or minimum values. Then, we use the second derivative test to check whether the critical point represents the minimum or maximum value of the function.

Complete step by step answer:
So, we have the function in x as ${x^2} - 3x + 5$. Let us assume the function ${x^2} - 3x + 5$ to be equal to y. So, $y = {x^2} - 3x + 5$. Now, we differentiate both sides of the above equation. So, we get,
$\Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {{x^2} - 3x + 5} \right)$
Separating the differentials of each term, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2}} \right) - \dfrac{d}{{dx}}\left( {3x} \right) + \dfrac{d}{{dx}}\left( 5 \right)$
Taking constants out of the differentials, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2}} \right) - 3\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( 5 \right)$
We know the power rule of differentiation as $\dfrac{{d\left[ {{x^n}} \right]}}{{dx}} = n{x^{\left( {n - 1} \right)}}$. So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = 2x - 3\left( 1 \right) + \left( 0 \right)$
We know that the derivative of a constant with respect to any variable is zero. Hence, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = 2x - 3 - - - - \left( 1 \right)$

Now, we equate the first derivative of the function to zero to find the critical points of the function. So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = 2x - 3 = 0$
So using the method of transposition, we will find the value of x.
$\Rightarrow 2x = 3$
$\Rightarrow x = \dfrac{3}{2}$
So, we get $x = \dfrac{3}{2}$ as a critical point of the function.
Now, we find the second derivative of the function. So, differentiating both sides of equation $\left( 1 \right)$ with respect to x, we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {2x - 3} \right)$
Separating the differentials,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {2x} \right) - \dfrac{d}{{dx}}\left( 3 \right)$

Again, using the power rule of differentiation, we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2\left( 1 \right) + 0$
So, we get the second derivative test as,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2$
Since the second derivative is positive for $x = \dfrac{3}{2}$. So, $x = \dfrac{3}{2}$ is the point of local minima. So, the minimum value of function is attained at $x = \dfrac{3}{2}$.
So, $y\left( {\dfrac{3}{2}} \right) = {\left( {\dfrac{3}{2}} \right)^2} - 3\left( {\dfrac{3}{2}} \right) + 5$
Computing the powers and simplifying the expression, we get,
$\Rightarrow y\left( {\dfrac{3}{2}} \right) = \dfrac{9}{4} - 3\left( {\dfrac{3}{2}} \right) + 5$

Opening the brackets, we get,
$\Rightarrow y\left( {\dfrac{3}{2}} \right) = \dfrac{9}{4} - \dfrac{9}{2} + 5$
Taking LCM, we get,
$\Rightarrow y\left( {\dfrac{3}{2}} \right) = \dfrac{9}{4} - \dfrac{{9 \times 2}}{{2 \times 2}} + \dfrac{{5 \times 4}}{{1 \times 4}}$
Simplifying further,
$\Rightarrow y\left( {\dfrac{3}{2}} \right) = \dfrac{9}{4} - \dfrac{{18}}{4} + \dfrac{{20}}{4}$
$\Rightarrow y\left( {\dfrac{3}{2}} \right) = \dfrac{{9 - 18 + 20}}{4}$
Adding in numerator,
$\therefore y\left( {\dfrac{3}{2}} \right) = \dfrac{{11}}{4}$

Therefore, the minimum value of the quadratic expression ${x^2} - 3x + 5$ is $\dfrac{{11}}{4}$.

Note: We can solve the problems involving the maxima and minima concept by two step methods: the first derivative test and the second derivative test. The first derivative test helps in finding the local extremum points of the function. The second derivative test involves finding the local extremum points and then finding out which of them is local minima and which is local maxima.