Question

The smallest value of ${x^2} - 3x + 3$ in the interval $\left( { - 3,\dfrac{3}{2}} \right)$ is
A. $\dfrac{3}{4}$
B. $5$
C. $ - 15$
D. $ - 20$

Answer 1

Hint: Use the concepts of minima and maxima which are applications of derivatives. Equate the above equation to $f(x)$ and find the first derivative of the function and equate it to 0 to find the critical point. Check whether you get the minimum value or the maximum value of the function at the critical point using the second derivative test.

Formula used: $\dfrac{{d({x^n} + l{x^m})}}{{dx}} = n{x^{n - 1}} + ml{x^{m - 1}}$

Complete step by step solution:
Let $f(x) = {x^2} - 3x + 3$.
Find the first derivative and equate it to 0 to find the critical point.
$f'(x) = 2x - 3 = 0$
$x = \dfrac{3}{2}$ is the critical point, c.
To check whether we get the minimum value or maximum value of $f(x)$ at c we need to check if $f''(c)$ is lesser than or greater than 0. If $f''(c) > 0$ then $f(c)$ is the minimum value of $f(x)$, if $f''(c) < 0$ then $f(c)$ is the maximum value of $f(x)$.
$f''(x) = 2$
Therefore, $f''(c) = 2 > 0$. $f(c)$ is the minimum value of $f(x)$.
$f\left( {\dfrac{3}{2}} \right) = {\left( {\dfrac{3}{2}} \right)^2} - 3\left( {\dfrac{3}{2}} \right) + 3 = \dfrac{9}{4} - \dfrac{9}{2} + 3$
$f\left( {\dfrac{3}{2}} \right) = \dfrac{3}{4}$
Therefore, the minimum value of ${x^2} - 3x + 3$ is $\dfrac{3}{4}$ and it occurs at $x = \dfrac{3}{2}$ which is in the interval $\left( { - 3,\dfrac{3}{2}} \right)$.
The correct answer is option A. $\dfrac{3}{4}$.

Note: Alternative method of solving:
${x^2} - 3x + 3 = {\left( {x - \dfrac{3}{2}} \right)^2} + \dfrac{3}{4}$
The minimum value cannot be lesser than $\dfrac{3}{4}$ because ${\left( {x - \dfrac{3}{2}} \right)^2}$ cannot be negative. Therefore, the smallest value of ${x^2} - 3x + 3$ in the interval $\left( { - 3,\dfrac{3}{2}} \right)$ is $\dfrac{3}{4}$.