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This is a bit confusing for me.

#Calculus

Note by Kushal Patankar
5 years, 10 months ago

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Comments

Try to realize that f(x)=x22x f(x) = x^2 - 2|x| . Try using the graph. If you still don't get, reply I'll add a detailed solution. Just to check are the answers C, D , D ?

Sudeep Salgia - 5 years, 10 months ago

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I am unable to graph g(x)g(x).

Kushal Patankar - 5 years, 10 months ago

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img img Here is the image for the graphs in the question. In the first part of g(x)g(x) , the value of g(x)g(x) is given by the minimum value of f f in the range (3,x) (-3 ,x) . Since the function is decreasing till 1 -1 , g(x)g(x) matches with f(x) f(x) . After that the value of g(x) g(x) becomes a constant equal to 1 -1 since that is the local minima. Similarly, we can plot it for the positive part. If it still sounds unclear, let me know I'll elaborate further. The answer to the third question should be C.

Sudeep Salgia - 5 years, 10 months ago

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@Sudeep Salgia Why did you graphed a straight line in (1,2)(1,2) , I was thinking of it to be f(x)f(x).

Kushal Patankar - 5 years, 10 months ago

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@Kushal Patankar It is maximum value of f(t)f(t) where 0tx 0 \leq t \leq x. Since the function is decreasing, the value would be f(0)f(0) which is 00.

Sudeep Salgia - 5 years, 10 months ago

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@Sudeep Salgia But in (1,2)(1,2) it is increasing. So all f(x)f(x) will be greater than f(t)f(t).

Kushal Patankar - 5 years, 10 months ago

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@Kushal Patankar Sorry for my misleading statement. It is true that the function is increasing in the interval (1,2)(1,2) but notice that the value is still negative, that is, f(r)<f(0)=0   r  (1,2) f(r) < f(0) = 0 \ \ \forall \ r \ \ \in (1,2) . Since t [0,x] t \ \in [0, x] , therefore  x<2,f(0)>f(x) \forall \ x < 2 , f(0) > f(x) .

Sudeep Salgia - 5 years, 10 months ago

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@Sudeep Salgia OK, got you, thanks ☺☺☺

Kushal Patankar - 5 years, 10 months ago

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@Kushal Patankar Happy to help. :)

Sudeep Salgia - 5 years, 10 months ago
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