Application of Derivatives — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives2 Marks
Question
Find the maximum and minimum value, f(x) = -(x - 1)2 + 10
✓
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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