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# In Problem with a problem

This is a bit confusing for me.

Note by Kushal Patankar
2 years, 1 month ago

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Try to realize that $$f(x) = x^2 - 2|x|$$. 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 ? · 2 years, 1 month ago

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I am unable to graph $$g(x)$$. · 2 years, 1 month ago

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img

Here is the image for the graphs in the question. In the first part of $$g(x)$$, the value of $$g(x)$$ is given by the minimum value of $$f$$ in the range $$(-3 ,x)$$. Since the function is decreasing till $$-1$$, $$g(x)$$ matches with $$f(x)$$. After that the value of $$g(x)$$ becomes a constant equal to $$-1$$ 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. · 2 years, 1 month ago

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Why did you graphed a straight line in $$(1,2)$$ , I was thinking of it to be $$f(x)$$. · 2 years, 1 month ago

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It is maximum value of $$f(t)$$ where $$0 \leq t \leq x$$. Since the function is decreasing, the value would be $$f(0)$$ which is $$0$$. · 2 years, 1 month ago

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But in $$(1,2)$$ it is increasing. So all $$f(x)$$ will be greater than $$f(t)$$. · 2 years, 1 month ago

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

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OK, got you, thanks ☺☺☺ · 2 years, 1 month ago

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Happy to help. :) · 2 years, 1 month ago

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