Given y=f(x), how many nature numbers n so that y=|f(x)+n| has 5 extremiums
#function math
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I got f(x) = x^4 - 2x^2 -3
heavy stump
Presumably by using f'(x) = a(x-1)(x+1)x expanding, integrating, and then using two points f(1) = -4 and f(0) = -3 to fix the constant a, and the constant of integration.
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@ripe crypt so when you abs, you fold the negative numbers into positive ones, right? This creates extrema at zeroes. So this function has 3 extrema normally, so as long as it crosses the x axis twice it will have two more extrema.
So this is equivalent to asking for how many natural numbers does this equation have only 2 zeroes.
If n = 0, this is true (is 0 a natural number for you?), and this remains true up until 3 where you have 3 zeroes, but it doesn't cross, so n=3 works.
heavy stump
I'm not sure what you mean by that
heavy stump
If n = 0, this is true (is 0 a natural number for you?), and this remains true u...
From here, you can test n=4 and find that there are two zeroes, but they are double zeroes, so they don't actually cross, so abs doesn't create new zeroes. So this still only has 3 extrema.
Beyond that the function has no zeroes.
heavy stump
Like homogeneous linear equations? Maybe that doesn’t apply
This is just a separable equation.
It's solvable with direct integration
heavy stump
Absolute value can create new extrema
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Nice
heavy stump
It can turn an odd counted zero into a minima.
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WOW
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From here, you can test n=4 and find that there are two zeroes, but they are dou...
Ohhh I got it
Tysm ❤️
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