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This is a bit confusing for me.

Note by Kushal Patankar
1 year, 12 months 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 ? · 1 year, 12 months ago

I am unable to graph $$g(x)$$. · 1 year, 12 months ago

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. · 1 year, 12 months ago

Why did you graphed a straight line in $$(1,2)$$ , I was thinking of it to be $$f(x)$$. · 1 year, 12 months ago

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 $$0$$. · 1 year, 12 months ago

But in $$(1,2)$$ it is increasing. So all $$f(x)$$ will be greater than $$f(t)$$. · 1 year, 12 months ago

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 $$\forall \ x < 2 , f(0) > f(x)$$ . · 1 year, 12 months ago

OK, got you, thanks ☺☺☺ · 1 year, 12 months ago