Question

If $m$ and $M$ respectively denote the minimum and maximum of $f\left(x\right)={\left(x-1\right)}^2+3$ for $x\in \left[-3,1\right]$, then the ordered pair $\left(m,M\right)$ is equal to

• A. $\left(-3,9\right)$
• B. $\left(3,19\right)$
• C. $\left(-19,3\right)$
• D. $\left(-19,-3\right)$

Answer 1

The correct option is B

$\left(3,19\right)$

Explanation for the correct option

Step 1: Solve for the value of $m$

The given function $f\left(x\right)={\left(x-1\right)}^2+3$.

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left[{\left(x-1\right)}^2+3\right] \\ \Rightarrow f\text{'}\left(x\right)=2\left(x-1\right)\frac{d}{dx}\left(x-1\right) \\ \Rightarrow f\text{'}\left(x\right)=2\left(x-1\right)\times 1 \\ \Rightarrow f\text{'}\left(x\right)=2x-2 \end{gathered}$$

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \Rightarrow \frac{d}{dx}\left[f\text{'}\left(x\right)\right]=\frac{d}{dx}\left[2x-2\right] \\ \Rightarrow f\text{'}\text{'}\left(x\right)=2 \end{gathered}$$

Thus, $f\text{'}\text{'}\left(x\right)>0$.

So, for the minimum value of $f\left(x\right)$, $f\text{'}\left(x\right)=0$

$$\begin{gathered} \Rightarrow 2x-2=0 \\ \Rightarrow x=1 \end{gathered}$$

Thus, the minimum value of $f\left(x\right)$, is $m=f\left(1\right)$.

$$\begin{gathered} \Rightarrow m={\left(1-1\right)}^2+3 \\ \Rightarrow m=3 \end{gathered}$$

Step 2: Solve for the value of $M$

The given domain of the function $f\left(x\right)$ is $x\in \left[-3,1\right]$.

Thus, the maximum value of $f\left(x\right)$ is $M=f\left(-3\right)$.

$$\begin{gathered} \Rightarrow M={\left(-3-1\right)}^2+3 \\ \Rightarrow M={\left(-4\right)}^2+3 \\ \Rightarrow M=16+3 \\ \Rightarrow M=19 \end{gathered}$$

Thus, the ordered pair $\left(m,M\right)$ is equal to $\left(3,19\right)$.

Hence, option(B) is the correct option.