Explain why f(x)=4x^7+7x^3+9x-357 does not have a tangent line with a slope of 1.

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Explain why f(x)=4x^7+7x^3+9x-357 does not have a tangent line with a slope of 1.

Mathematics
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we can take the derivative of it and try to make it equal 1
f(x)= 4x^7+7x^3+9x-357 f'(x)= 28x^6+21x^2+9 = 1 28x^6 +21x^2 +8 = 0 has the same effect
we can use synthetic division and trial and error to try to come to the roots...

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Other answers:

1, 2, 4, 7, 14, 28 --------------- these numbers will give us a "pool" of options 1, 2, 4, 8 that we can use to determine its roots
then its just a matter of digging in and getting dirty :)
i was told that the answer has to do with the degree of the polynomial (the derivative polynomial the 28x^6+21x^2+9)
im sure it does, but without any prior knowledge about any shortcuts, this is how I would go about convincing someone that there is no slope of 1
there are therums with names that I dont remember that tell you if the signs are all positive, then there is no positive root....
lets try x = 1 on this... -------------------- 1 | 28 0 0 0 21 0 9 0 28 ........ 28......49 -------------------- 28 28 ......... 49......58 since all our answers are positive, then (x-1) and higher are not going to be answers
\[x^{2n} \geq 0\ \forall{ x \in \mathbb{R}; n \in \mathbb{Z}}.\] Therefore it is > 0 ...
Sorry, the LaTeX engine on this thing is pellet so those conditions may be a little odd looking
since 1 is not a root, it leads me to believe that there is no tangent to the curve that has a slope of 1....
looks greek to me :)
x^{even} >= 0 for all x
that "A" means all x elements of R right?
So it + 9 > 1 for all x
"for all" yes
well the polynomial does not even hits the x-axis so there can't be any roots for the polynomial i guess; they're all imaginary roots
if the graph of the derivative doesnt hit the x axis, then right, there are no real root...
Tangent with gradient 1 iff f'(x) = 1 iff 28x^6+21x^2+9 = 1 iff 28x^6+21x^2 + 8 = 0 , from above. And I have just told you that the first part is always >= 0 .
Yes, listen to amistre
ok, well thanks to both of you
youre welcome :)

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