Find K for which f is one-to-one

  • Thread starter intenzxboi
  • Start date
In summary, the function f(x)= x^3 + kx^2 + 19x is one-to-one if k is between -sqrt(57) and sqrt(57). At k=0, the function is increasing and at k=+/-3, the function is decreasing. At k=+/-sqrt(57), the function is neither increasing nor decreasing and therefore not one-to-one.
  • #1
intenzxboi
98
0

Homework Statement



f(x)= x^3 + kx^2 + 19x


The Attempt at a Solution



So what i did was take the derivative of f(x)
3x^2 + 2kx + 19
\sqrt{}


(3x+ [tex]\sqrt{19}[/tex]) (x+ [tex]\sqrt{19}[/tex])

so then i can assume that 3x[tex]\sqrt{19}[/tex] + x[tex]\sqrt{19}[/tex] = 2kx

and then k i got to be 2[tex]\sqrt{19}[/tex]

Im i right?? i feel like I'm completely lost..
 
Physics news on Phys.org
  • #2
f(x) is one to one if it's derivative never changes sign. f'(x)=3x^2+2kx+19 is always positive for large enough x, so you want to find the values k where f'(x) is never negative. Think about finding the minimum of f'(x) as a function of k.
 
  • #3
k so find where f(x) is increasing

so..

3x + [tex]\sqrt{19}[/tex] > 0

x + [tex]\sqrt{19}[/tex] > 0

So x> (-[tex]\sqrt{19}[/tex]) / 3
and x> -[tex]\sqrt{19}[/tex]

Is that right??
 
  • #4
intenzxboi said:

Homework Statement



f(x)= x^3 + kx^2 + 19x


The Attempt at a Solution



So what i did was take the derivative of f(x)
3x^2 + 2kx + 19
\sqrt{}


(3x+ [tex]\sqrt{19}[/tex]) (x+ [tex]\sqrt{19}[/tex])
What happened to k? It shouldn't just go away.
intenzxboi said:
so then i can assume that 3x[tex]\sqrt{19}[/tex] + x[tex]\sqrt{19}[/tex] = 2kx

and then k i got to be 2[tex]\sqrt{19}[/tex]

Im i right?? i feel like I'm completely lost..
 
  • #5
intenzxboi said:
k so find where f(x) is increasing

so..

3x + [tex]\sqrt{19}[/tex] > 0

x + [tex]\sqrt{19}[/tex] > 0

So x> (-[tex]\sqrt{19}[/tex]) / 3
and x> -[tex]\sqrt{19}[/tex]

Is that right??
No.
Go back to the derivative you found and determine a value of k that makes the derivative positive everywhere. Instead of factoring, as you have attempted (unsuccessfully) to do, you might try the quadratic formula.
 
  • #6
k i used the quadratic formula and got

[ -2k +/- [tex]\sqrt{2k^2 - 228}[/tex] ] / 6 = X

Having trouble solving for k
 
  • #7
No, no, no. Find the minimum of f'(x). Take the derivative of f'(x) (i.e. f''(x)) and set it equal to zero. That will give you the x value of the point where f'(x) is a minimum. What is the value of f'(x) there?
 
  • #8
You have f'(x) = 3x^2 + 2kx + 19, and you would like to find a value of k so that f'(x) > 0 for all x.

By using the Quadratic formula, so solved the equation 3x^2 + 2kx + 19 = 0, for x.
If you want 3x^2 + 2kx + 19 to not have any (real) solutions for x, you want to pick a number k so that the discriminant (the part in the radical in your work) is negative. If that happens, there won't be any real solutions to the equation 3x^2 + 2kx + 19 = 0.

What does it take for the discriminant to be negative?
 
  • #9
Taking the 2nd derivative i get 6x + 2k = 0
so x=-2k/6


take this values into the 1st derivative and i get a long equation that ends up giving me after reducing

-k^2 - 2k/3 + 19
 
  • #10
You have that the minimum of f'(x) is at x=-k/3. That's good. But when I put x=-k/3 into 3x^2 + 2kx + 19 I DON'T get -k^2 - 2k/3 + 19. Could you fix that up?
 
  • #11
Opps i forgot the squared root sign.

-k^2 - (2k^2)/3 + 19

Do i set this to 0?
 
  • #12
intenzxboi said:
Opps i forgot the squared root sign.

-k^2 - (2k^2)/3 + 19

Do i set this to 0?

I don't agree with the first term either. It was supposed to be 3*(-k/3)^2. Sure, once you get it right, then you can set it equal to zero and use that info to figure out where it's nonnegative.
 
  • #13
arg... okay i for some reason cancel the 3 out..

new equation gave me k^2= 19
 
  • #14
Your algebra isn't doing so well today. 3*(-k/3)^2=k^2/3. Plus. Not minus.
 
  • #15
yea i think i got it.. half way thru the equation i figures 3x root19 + x root19
need a minus sign
i ma start all over.. give me a sec
 
  • #16
intenzxboi said:
yea i think i got it.. half way thru the equation i figures 3x root19 + x root19
need a minus sign
i ma start all over.. give me a sec

Don't start all over! You just need a good expression for the minimum of f'(x) in terms of k. Just fix the sign problem.
 
  • #17
No no i save the information on like 20 pages.. wanted to do it on one neat page for my records later on.

I finally got k= +/- root 57
 
  • #18
Right. Those are the values of k for which f'(x) has a zero minimum. The minimum in general is 19-k^2/3. For what values of k is that nonnegative? +/-sqrt(57) aren't the only possible values of k.
 
  • #19
Thank You!

I finally understand it

so if I am correct

the function x^3 + kx^2 + 19x

is one to one at
k <= - root 57 OR root 57 <= k
 
  • #20
Other way around. -sqrt(57)<=k<=sqrt(57), right? For example, k=0 works.
 
  • #21
Huh??
k i have

For increasing
-k^2 /3 >-19

k< +/- root 57

For decreasing
-k^2 /3 <-19
k> +/- root 57So it is a one-to-one if its either increasing or decreasing
 
  • #22
You want to find the values where the minimum of the first derivative is nonnegative, right? Agreed? You found that the minimum of f'(x) is 19-k^2/3, ok so far? The minimum is zero at k=+/-sqrt(57)~+/-7.5. But at k=0, the minimum is 19. At k=+/-3 the minimum is 16, ok, still positive. At k=+/-10 the minimum is negative. So for say k=10, the function f(x) increases, then decreases then increases. It's not one to one. Correct?
 

Related to Find K for which f is one-to-one

1. How do you determine if a function is one-to-one?

A function is one-to-one when each input has a unique output. This means that no two inputs can have the same output. To determine if a function is one-to-one, you can use the horizontal line test. If a horizontal line intersects the graph of the function at more than one point, then the function is not one-to-one.

2. Why is it important to find K for which f is one-to-one?

Finding K for which f is one-to-one is important because it helps ensure that the function has an inverse. A one-to-one function has a unique inverse, which allows us to find the original input (x) from the output (y) by using the inverse function.

3. How can you find K for which f is one-to-one?

To find K for which f is one-to-one, you can set up an equation where the function is equal to a variable, such as y = Kx. Then, you can use algebraic techniques to solve for K. Once you have found the value of K, you can substitute it back into the original function to confirm that it is one-to-one.

4. Can a function be one-to-one for some values of K but not others?

Yes, a function can be one-to-one for some values of K but not others. For example, the function f(x) = x^2 is not one-to-one for all values of K, but if we restrict the domain to only positive values, then the function becomes one-to-one.

5. How does finding K for which f is one-to-one affect the domain and range of the function?

Finding K for which f is one-to-one can affect the domain and range of the function. If a function is not one-to-one, there may be some inputs that do not have an output, which would affect the domain. Additionally, if the function is not one-to-one, there may be some outputs that have multiple inputs, which would affect the range. By finding K for which f is one-to-one, we can ensure that the function has a well-defined domain and range.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
665
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
25
Views
665
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
877
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
904
  • Calculus and Beyond Homework Help
Replies
16
Views
2K
Back
Top