Intuitively, looking at the graph, the maximum will be when the slopes of the two lines are equal. To one side, one line is rising faster than the other and the distance is either growing or shrinking. So the slope of y=x+42 is 1 all the time, where is the slope of y=x^2 equal to 1? Obviously the f(y)` = 2x, so 1= 2x => x= 1/2. Kind of a 'cheat', but this problem lends itself to this way of looking at it.
from what i've gathered he uses y2-y1 because he is calculating VERTICAL distance. that distance formula is better used when you have to calculate the distance between 2 known points (especially when horizontal and vertical distance is counted for)
Thank you sooo much for explaining the reason behind each steps. Explained very clearly.
Excellent explanation Sir. Thanks 🙏
Interesting and very helpful. Thank u sir
This video is very interesting
super well explained thanks!
Intuitively, looking at the graph, the maximum will be when the slopes of the two lines are equal. To one side, one line is rising faster than the other and the distance is either growing or shrinking. So the slope of y=x+42 is 1 all the time, where is the slope of y=x^2 equal to 1? Obviously the f(y)` = 2x, so 1= 2x => x= 1/2. Kind of a 'cheat', but this problem lends itself to this way of looking at it.
I cleared my doubt
Sir l have learned that the distance between x1y1 and x2y2 is [(x2-x1)^2 +(y2-y1)^2]^1/2 .but your method is different.why?
Use the method you know
from what i've gathered he uses y2-y1 because he is calculating VERTICAL distance. that distance formula is better used when you have to calculate the distance between 2 known points (especially when horizontal and vertical distance is counted for)
woah man thats tough, even your subscribers aren't subscribed to you *sad_emoji*