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Maxima and Minima — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMaxima and Minima5 Marks
Question
Find the absolute maximum and the absolute minimum value of the following functions in the given intervals:
$f(x) = (x - 1)^2 + 3$ in $[-3, 1]$
✓

Answer

The given function is $f(x) = (x - 1)^2+ 3.$
$\therefore f'(x) = 2(x - 1)$
Now, $f'(x) = 0 $
$\Rightarrow 2(x - 1) = 0 $
$\Rightarrow x = 1$
Then, we evalute the value of $f$ at critical point $x = 1$ and at the end point of the interval $[-3, 1].$
$f(1) = (1 - 1)^2 + 3 = 0 + 3 = 3$
$f(-3) = (-3 - 1)^2 + 3 = 16 + 3 = 19$
Hence, we can conclude that the absolute maximum value of f on $[-3, 1]$ is occurring at $x = -3$ and the minimum value of f on $[-3, 1]$ is $3$ occurring at $x = 1.$

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