Find K for which f is one-to-one

[intenzxboi]

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+ $$\sqrt{19}$$) (x+ $$\sqrt{19}$$)

so then i can assume that 3x$$\sqrt{19}$$ + x$$\sqrt{19}$$ = 2kx

and then k i got to be 2$$\sqrt{19}$$

Im i right?? i feel like I'm completely lost..

 

[Dick]

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.

 

[intenzxboi]

k so find where f(x) is increasing

so..

3x + $$\sqrt{19}$$ > 0

x + $$\sqrt{19}$$ > 0

So x> (-$$\sqrt{19}$$) / 3
and x> -$$\sqrt{19}$$

Is that right??

 

[Mark44]

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+ $$\sqrt{19}$$) (x+ $$\sqrt{19}$$)

What happened to k? It shouldn't just go away.

so then i can assume that 3x$$\sqrt{19}$$ + x$$\sqrt{19}$$ = 2kx

and then k i got to be 2$$\sqrt{19}$$

Im i right?? i feel like I'm completely lost..

 

[Mark44]

k so find where f(x) is increasing

so..

3x + $$\sqrt{19}$$ > 0

x + $$\sqrt{19}$$ > 0

So x> (-$$\sqrt{19}$$) / 3
and x> -$$\sqrt{19}$$

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.

 

[intenzxboi]

k i used the quadratic formula and got

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

Having trouble solving for k

 

[Dick]

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?

 

[Mark44]

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?

 

[intenzxboi]

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

 

[Dick]

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?

 

[intenzxboi]

Opps i forgot the squared root sign.

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

Do i set this to 0?

 

[Dick]

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.

 

[intenzxboi]

arg... okay i for some reason cancel the 3 out..

new equation gave me k^2= 19

 

[Dick]

Your algebra isn't doing so well today. 3*(-k/3)^2=k^2/3. Plus. Not minus.

 

[intenzxboi]

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

 

[Dick]

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.

 

[intenzxboi]

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

 

[Dick]

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.

 

[intenzxboi]

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

 

[Dick]

Other way around. -sqrt(57)<=k<=sqrt(57), right? For example, k=0 works.

 

[intenzxboi]

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

 

[Dick]

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?