This is a perfect example where an always increasing function intersects an always decreasing function. The key difference is that for it to have only 1 solution it must be continuously decreasing but because it is discontinuous at x=0 it’s decreasing then picks back up and decreases again breaking the continuity.
I took a little different more tedious rigorous approach I started to look at the problem from an integer mindset because I had a feeling dealing with rational numbers would be like hell I brought the -5x to the R.H.S x(3^x+1)-5x=4 factor out the x ----> x(3^x+1 -5)=4 as i was dealing with integers, x must divide 4 therefore possible x values are:1,-1,2,-2,4,-4 Then when you plug those values in you get two integer solutions x=1 and x=-1
The analysis of f'(x) is wrong. f'(x)=3^(x+1) + x×ln(3)×3^(x+1). In the video, you left out the factor of x. In particular, f'(x)=0 when x=-1/ln(e). (That's the only zero.)
Then we could calculate an estimation of each solution by numerical iteration. Or if we're lucky, we could easily see the non-rational solution, or perform a substitution that makes the non-rational solution easy to see.
Upon inspection, it can be easily seen that x=0 is not a solution, and x=1 is a solution (both sides of the equation become 9). The question now is, if x=1 is the only solution. Let's rewrite the equation: x * 3^(x+1) = 5x + 4 x * 3^(x+1) - 5x = 4 ... since x=0 is not a solution, we can safely divide both sides by x ... 3^(x+1) - 5 = 4/x By sketching the graphs of f(x) = (3^(x+1) - 5) and g(x) = 4/x in one plot, it is easily seen that f(x) and g(x) have two intersection points: one intersection for positive x , and one intersection for negative x . There is only one intersection for positive x, because for positive x, f(x) is continuous and strictly increasing while g(x) is continuous and strictly decreasing. And the same argument explains why there is only one intersection for negative x. The positive solution has already been identified (namely x=1), the negative solution is easily found at x = -1 (both sides of the original equation become -1 ). So the solutions are x = 1 and x = -1 .
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2:30 I think you missed an x in the second part of the derivative.
Yes, for sure he missed the x !! 👏👏
Luv a graph
This is a perfect example where an always increasing function intersects an always decreasing function. The key difference is that for it to have only 1 solution it must be continuously decreasing but because it is discontinuous at x=0 it’s decreasing then picks back up and decreases again breaking the continuity.
Good point!
I took a little different more tedious rigorous approach
I started to look at the problem from an integer mindset because I had a feeling dealing with rational numbers would be like hell
I brought the -5x to the R.H.S
x(3^x+1)-5x=4 factor out the x ----> x(3^x+1 -5)=4
as i was dealing with integers, x must divide 4 therefore possible x values are:1,-1,2,-2,4,-4
Then when you plug those values in you get two integer solutions x=1 and x=-1
The analysis of f'(x) is wrong. f'(x)=3^(x+1) + x×ln(3)×3^(x+1). In the video, you left out the factor of x. In particular, f'(x)=0 when x=-1/ln(e). (That's the only zero.)
My first thought was, we need to check the polynomial factors for x=1 and x=-1. Now how do we adapt for this equation?
This is great. I am wondering how we would find the solutions if the solutions were not rational numbers.
Then we could calculate an estimation of each solution by numerical iteration. Or if we're lucky, we could easily see the non-rational solution, or perform a substitution that makes the non-rational solution easy to see.
Upon inspection, it can be easily seen that x=0 is not a solution, and x=1 is a solution (both sides of the equation become 9). The question now is, if x=1 is the only solution. Let's rewrite the equation:
x * 3^(x+1) = 5x + 4
x * 3^(x+1) - 5x = 4
... since x=0 is not a solution, we can safely divide both sides by x ...
3^(x+1) - 5 = 4/x
By sketching the graphs of f(x) = (3^(x+1) - 5) and g(x) = 4/x in one plot, it is easily seen that f(x) and g(x) have two intersection points: one intersection for positive x , and one intersection for negative x . There is only one intersection for positive x, because for positive x, f(x) is continuous and strictly increasing while g(x) is continuous and strictly decreasing. And the same argument explains why there is only one intersection for negative x.
The positive solution has already been identified (namely x=1), the negative solution is easily found at x = -1 (both sides of the original equation become -1 ).
So the solutions are x = 1 and x = -1 .
[Kinda first but never mind] Could this be solved by calculus ? Anyways that's such a great interesting video, love your content !
x = 1
I got 1 but not -1.