Question

question

Answer 1

are you asking who has a question

Answer 2

no he is gonna post lol

Answer 3

she is* my bad

Answer 4

\(\large \color{black}{\begin{align}

& \normalsize \text{Let }\ a\in \mathbb{R} \ \normalsize \text{and let }\ f: \mathbb{R}\rightarrow \mathbb{R} \normalsize \text{be given by } \hspace{.33em}\\~\\
& f(x)=x^5-5x+a \hspace{.33em}\\~\\
& \normalsize \text{then } \hspace{.33em}\\~\\
& A.) \quad f(x)\ \normalsize \text{has three real roots if }\ a>4 \hspace{.33em}\\~\\
& B.) \quad f(x)\ \normalsize\text{has only one real roots if }\ a>4 \hspace{.33em}\\~\\
& C.) \quad f(x)\ \normalsize\text{has only three real roots if }\ a<-4 \hspace{.33em}\\~\\
& D.) \quad f(x)\ \normalsize\text{has only three real roots if }\ -4<a<4 \hspace{.33em}\\~\\
\end{align}}\)

Answer 5

hmm what did you try?

Answer 6

This info might be useful: https://en.wikipedia.org/wiki/Quintic_function#Quintics_in_Bring.E2.80.93Jerrard_form

Answer 7

putting a=5

Answer 8

honestly this question is beyond my IQ

Answer 9

why a=5
you are trying to find the constraints for a

Answer 10

for first 2 options it says to check roots for a>4

Answer 11

oh i see your line of thinking

Answer 12

answer given is both option (B.) and (D.)

Answer 13

hmm not sure if this might help
suppose that has 3 distinct real roots
therefore
\[f(x)=x^5-5x+a=(x-l)(x-h)(x-r)(Ax^2+Bx+C)\]
where Ax^2+Bx+C has no real roots

Answer 14

and how do i factor that

Answer 15

Long divide by a factor (x-root)

Answer 16

via the Fund. thm of algebra

Answer 17

i don't even know what thm is
but i think this needs some algebra thm application

Answer 18

i just need a value for \(a\) so that the expression is factorrable

Answer 19

for all constraints of \(a\) as given in options

Answer 20

you mean you want to go trial and error ?

Answer 21

yes

Answer 22

if it is the shortest soln

Answer 23

this is doable by calculus if we look for f"
but you didn't take that yet

Answer 24

@freckles any bright ideas

Answer 25

let's google haha

Answer 26

haha lol

Answer 27

can we use descartus rule of signs here

Answer 28

Hey @mathmath333 you can't do derivatives right?

Answer 29

or calculus stuff?

Answer 30

i believe that rule won't help here

Answer 31

not yet at least

Answer 32

well i cannt but if u can use simple calculus then i can understand

Answer 33

example sith and giggles used calculus here ,that level i can understand

http://openstudy.com/users/mathmath333#/updates/55818ea7e4b0636b8cc4098d

Answer 34

from here this confirms what you said B and D

Answer 35

https://www.desmos.com/calculator/dsmahabxqt

Answer 36

\[f(x)=x^5-5x+a \\ f'(x)=5x^4-5 \\ f'(x)=0 \text{ when } 5x^4-5=0 \\ 5x^4-5=0 \\ 5(x^4-1)=0 \\ (x^2-1)(x^2+1)=0 \\ x^2-1=0 \\ x=\pm 1 \\ f(1)=1-5+a \text{ and } f(-1)=-1+5+a \\ f(1)=a-4 \text{ and } f(-1)=a+4 \]
I think we should play with the signs of f(-1) and f(1)

Answer 37

why did u took first derivative and equated it to zero

Answer 38

that's to find maxima and minima

Answer 39

y we need maxima and minima

Answer 40

at first I was looking at cases for the above and deleted all of that because it was totally unnecessary after playing with the cases and a little calculus on the cases
to get a better sense of the graph)
and where the function is increasing or decreasing if at all
\[f'(x)=5x^4-5=5(x^4-1)=5(x^2-1)(x^2+1) \\ 5(x^2-1)(x^2+1)>0 \\ x^2-1>0 \\ x^2>1 \text{ so increasing on } x \in (-\infty,1) \cup (1,\infty)\]
\[\text{ so decreasing on } x \in (-1,1)\]


anyways if both f(1) and f(-1) are positive then your graph looks like this:



and if we have both f(1) and f(-1) are negative then your graph looks like:



now if f(1) is negative and f(-1) is positive your graph looks like this: