This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
When posting on Brilliant:
Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
Markdown
Appears as
*italics* or _italics_
italics
**bold** or __bold__
bold
- bulleted - list
bulleted
list
1. numbered 2. list
numbered
list
Note: you must add a full line of space before and after lists for them to show up correctly
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
Math
Appears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
Try to realize that f(x)=x2−2∣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 ?
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.
@Kushal Patankar
–
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∀r∈(1,2). Since t∈[0,x], therefore ∀x<2,f(0)>f(x) .
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Try to realize that f(x)=x2−2∣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 ?
Log in to reply
I am unable to graph g(x).
Log in to reply
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.
Log in to reply
(1,2) , I was thinking of it to be f(x).
Why did you graphed a straight line inLog in to reply
f(t) where 0≤t≤x. Since the function is decreasing, the value would be f(0) which is 0.
It is maximum value ofLog in to reply
(1,2) it is increasing. So all f(x) will be greater than f(t).
But inLog in to reply
(1,2) but notice that the value is still negative, that is, f(r)<f(0)=0 ∀ r ∈(1,2). Since t ∈[0,x], therefore ∀ x<2,f(0)>f(x) .
Sorry for my misleading statement. It is true that the function is increasing in the intervalLog in to reply
Log in to reply