Find the maximum and minimum value, f(x) = -(x - 1)^2 + 10 - Vidyadip

Question

Find the maximum and minimum value, $f(x) = -(x - 1)^2 + 10$

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Answer

It is given that $f(x) = -(x - 1)^2 + 10$
Now, we can see that $(x - 1)^2 \ge 0$ for every $x \in R$
$\Rightarrow f(x) = -(x - 1)^2 + 10 \le 10$ for every $x \in R$
The maximum value of $f$ is attained when $x - 1 = 0$
i.e, $x - 1 = 0$
$\Rightarrow x = 1$
Then, Maximum value of $f = f(1) = -(1 - 1)^2 + 10 = 10$
Since, $x = 1$ is the only critical point,
Therefore, function $f$ does not have a minimum value.

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