Question

Please help!
1) Show that the function y=x^21+x^11+13x does not have local maximum or local minimum.
2) Find the values of a and b if the function y=2x^3+ax^2+bx+36 has local maximum when x=-4 and a local minimum when x=5.
Thanks! :)

Answer 1

Mmmm ok, remember how to find max/min (critical) points?

Answer 2

Yes

Answer 3

So our first derivative tells us about slope.
Setting it equal to 0 gives us zero slopes, critical points.
Let's try to do that.

Answer 4

\[\Large\rm y=x^{21}+x^{11}+13x\]\[\Large\rm y'=?\]

Answer 5

\[y'=21x ^{20}+11x ^{10}+13\]

Answer 6

Setting equal to zero gives us,\[\Large\rm 0=21x^{20}+11x^{10}+13\]Which I'm going to rewrite like this,\[\Large\rm 0=21(x^{10})^2+11(x^{10})+13\]That step ok?
I applied an exponent rule there.

Answer 7

Yep

Answer 8

I'll make the substitution \(\Large\rm u=x^{10}\),\[\Large\rm 0=21u^2+11u+13\]

Answer 9

If we throw this into the quadratic equation we'll find that the discriminant is less than zero (The part under the root), so no real solutions exists for \(\Large\rm u\). Which further implies that no real solutions exists for \(\Large\rm x\).

Answer 10

That shows us that no critical points exist for this function.
Therefore no max/min points will exist.

Answer 11

Any of that confusing?
Do you remember what the quadratic equation looks like?

Answer 12

Yes, I think i got it! Thanks for the great explanation.

Answer 13

Mmmm how bout part 2?

Answer 14

Yes, please!

Answer 15

So the process starts out the same way, yah?
What do you get for your derivative?

Answer 16

\[y'= 6x^2+2ax+b\] ???

Answer 17

\[\Large\rm y=2x^3+ax^2+bx+36\]\[\Large\rm y'=6x^2+2ax+b\]Mmmm ok looks good!

Answer 18

They gave us the exact locations of max and min points of this function.
Since \(\Large\rm x=5\) is a maximum of the function,
it is also a critical point of the function,
and since it is a critical point, it will result in the first derivative equaling zero.

Answer 19

So plugging in y'=0 and x=5 gives us,\[\Large\rm 0=6(5)^2+2a(5)+b\]

Answer 20

Since x=5 is a minimum*
blah typo, still works out ok though

Answer 21

okay..

Answer 22

This gives us an equation involving a, b, and numbers.
We want to do the same thing with the x=-4 information,

this will give us a `system of 2 equations and 2 unknowns`.

Answer 23

So this first equation we can write as:\[\Large\rm -150=10a+b\]

Answer 24

How bout the other one, can you figure that equation out? :)

Answer 25

Is it -96=-8a+b ?

Answer 26

Mmm ok very good!

Answer 27

So we have a system of 2 equations, 2 unknowns.

Answer 28

Do you remember how to solve a system like that?

Answer 29

I think I forgot how to do it...

Answer 30

We would like to combine these equations in some way that causes one of the variables to disappear.