Prove that $x^7-2x^5+10x^2-1$ has no root greater than 1.
In the previous blog post, I mentioned of solving the equation $x^7-2x^5+10x^2-1=0$ so to show that the function $x^7-2x^5+10x^2-1$ has no root greater than 1.
But, that is a really bad idea. The reason why the question setters stated the problem so mostly because they wanted to avoid us to solve for the problem. To solve for the polynomial of degree seven is really difficult, plus, the polynomial couldn't be factorized and so the real roots are kind of "ugly", there are no exact values for them.
So, we ditch that idea for good.
Now what? Looking at the graph graph of y=x^7-2x^5+10x^2-1 again, what can you tell the part of the graph where $x\gt 1$?
What behavior of the graph could you observe from it?
Oh yes! It's strictly increasing at $x\gt 1$. Hurray! As long as we can prove the function is strictly increasing in the interval $x\gt 1$, and at $x=1$, $x^7-2x^5+10x^2-1=1-2+10-1=8\ne 0$, those mean the curve of the graph won't intersect the $x$ axis at $x\gt 1$.
And speaking of strictly increasing, we know we must rely on calculus method to prove the function $x^7-2x^5+10x^2-1$ is strictly increasing for $x\gt 1$.
We first let $f(x)=x^7-2x^5+10x^2-1$, then, we look for its first derivative. If we can prove the first derivative is always greater than zero in the interval $x\gt 1$, then we're done.
$f(x)=x^7-2x^5+10x^2-1$
$f'(x)=7x^6-10x^4+20x$
The second problem that arises now is how are we going to prove $f'(x)=7x^6-10x^4+20x\gt 0$? Again, the expression $7x^6-10x^4+20x$ couldn't be factored, and we have hit to a plateau now...
I wish to hear from you how you're going to prove $f'(x)=7x^6-10x^4+20x\gt 0$, because if you tried hard enough on any particular problem without fruitful solution, then, when the answer is revealed, you would kick yourself (in a good way) why didn't you think of it, most importantly, the solution will appear so appealing to you that the concept would retain in your mind forever.
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