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

Question

amistre64 · Answer

we can take the derivative of it and try to make it equal 1

amistre64 · Answer

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

amistre64 · Answer

we can use synthetic division and trial and error to try to come to the roots...

amistre64 · Answer

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

amistre64 · Answer

then its just a matter of digging in and getting dirty :)

anonymous · Answer

i was told that the answer has to do with the degree of the polynomial (the derivative polynomial the 28x^6+21x^2+9)

amistre64 · Answer

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

amistre64 · Answer

there are therums with names that I dont remember that tell you if the signs are all positive, then there is no positive root....

amistre64 · Answer

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

anonymous · Answer

$$x^{2n} \geq 0\ \forall{ x \in \mathbb{R}; n \in \mathbb{Z}}.$$

Therefore it is > 0  ...

anonymous · Answer

Sorry, the LaTeX engine on this thing is pellet so those conditions may be a little odd looking

amistre64 · Answer

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....

amistre64 · Answer

looks greek to me :)

anonymous · Answer

x^{even} >= 0 for all x

amistre64 · Answer

that "A" means all x elements of R right?

anonymous · Answer

So it + 9 > 1 for all x

anonymous · Answer

"for all" yes

anonymous · Answer

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

amistre64 · Answer

if the graph of the derivative doesnt hit the x axis, then right, there are no real root...

anonymous · Answer

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 .

anonymous · Answer

Yes, listen to amistre

anonymous · Answer

ok, well thanks to both of you

amistre64 · Answer

youre welcome :)