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If $m$ and $M$ respectively denote the minimum and maximum of $f(x)=(x-1)^2+3$ for $x \in[-3,1]$, then the ordered pair $(m, M)$ is equal to
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Verified Answer
The correct answer is:
$(3,19)$
Given that,
$$
f(x)=(x-1)^2+3, \quad x \in[-3,1]
$$
On differentiating w.r.t. $x$, we get
$$
f^{\prime}(x)=2(x-1)
$$
For maxima and minima, put $f^{\prime}(x)=0$
$$
\begin{aligned}
\Rightarrow & & 2(x-1) & =0 \\
\Rightarrow & & x & =1
\end{aligned}
$$
Now, $f^{\prime \prime}(x)=2$, minima $\forall x \in R$
At $\quad x=1$,
$$
f(1)=(1-1)^2+3=3
$$
At
$$
\begin{aligned}
x & =-3, \\
f(-3) & =(-3-1)^2+3=19
\end{aligned}
$$
Here, $m=3$ and $M=19$
Hence, required ordered pair is $(3,19)$
$$
f(x)=(x-1)^2+3, \quad x \in[-3,1]
$$
On differentiating w.r.t. $x$, we get
$$
f^{\prime}(x)=2(x-1)
$$
For maxima and minima, put $f^{\prime}(x)=0$
$$
\begin{aligned}
\Rightarrow & & 2(x-1) & =0 \\
\Rightarrow & & x & =1
\end{aligned}
$$
Now, $f^{\prime \prime}(x)=2$, minima $\forall x \in R$
At $\quad x=1$,
$$
f(1)=(1-1)^2+3=3
$$
At
$$
\begin{aligned}
x & =-3, \\
f(-3) & =(-3-1)^2+3=19
\end{aligned}
$$
Here, $m=3$ and $M=19$
Hence, required ordered pair is $(3,19)$
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