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6. Application of Derivatives
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Q1 of 168 Page 231

Find the maximum and minimum values, if any, of the following functions given by

f(x) = – (x – 1)2 + 10

It is given that f (x) = –(x – 1)2 + 10

Now, we can see that (x - 1)2 ≥ 0 for every x ϵ R


⇒ f (x) = –(x – 1)2 + 10 ≤ 10 for every x ϵ R


The minimum value of f is attained when x - 1 = 0


x - 1 = 0


⇒ x = 1


Then, Maximum value of f = f(1) = -(1-1)2 + 10 = 10


Therefore, function f does not have a minimum value.


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Questions · 168
6. Application of Derivatives
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