anonymous
  • anonymous
Explain why f(x)=4x^7+7x^3+9x-357 does not have a tangent line with a slope of 1.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
we can take the derivative of it and try to make it equal 1
amistre64
  • amistre64
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
  • amistre64
we can use synthetic division and trial and error to try to come to the roots...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
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
  • amistre64
then its just a matter of digging in and getting dirty :)
anonymous
  • anonymous
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
  • amistre64
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
  • amistre64
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
  • amistre64
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
  • anonymous
\[x^{2n} \geq 0\ \forall{ x \in \mathbb{R}; n \in \mathbb{Z}}.\] Therefore it is > 0 ...
anonymous
  • anonymous
Sorry, the LaTeX engine on this thing is pellet so those conditions may be a little odd looking
amistre64
  • amistre64
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
  • amistre64
looks greek to me :)
anonymous
  • anonymous
x^{even} >= 0 for all x
amistre64
  • amistre64
that "A" means all x elements of R right?
anonymous
  • anonymous
So it + 9 > 1 for all x
anonymous
  • anonymous
"for all" yes
anonymous
  • anonymous
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
  • amistre64
if the graph of the derivative doesnt hit the x axis, then right, there are no real root...
anonymous
  • anonymous
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
  • anonymous
Yes, listen to amistre
anonymous
  • anonymous
ok, well thanks to both of you
amistre64
  • amistre64
youre welcome :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.