increasing function

[bhuvaneshnick]

EM 2.39 . .If \(\displaystyle f(x)\, =\, kx^3\, -\, 9x^2\, +\, 9x\, +\, 3\) is increasing in each interval, then:

. . . . . . . . . .(A) \(\displaystyle k\, <\, 3\) . . . (B) \(\displaystyle k\, \leq\, 3\) . . . (C) \(\displaystyle k\, >\, 3\) . . . (D) \(\displaystyle k\, \geq\, 3\)

I don't understand the question. could some one explain me this. I am messing up with "in each interval"? Thank you

 

[Ishuda]

EM 2.39 . .If \(\displaystyle f(x)\, =\, kx^3\, -\, 9x^2\, +\, 9x\, +\, 3\) is increasing in each interval, then:

. . . . . . . . . .(A) \(\displaystyle k\, <\, 3\) . . . (B) \(\displaystyle k\, \leq\, 3\) . . . (C) \(\displaystyle k\, >\, 3\) . . . (D) \(\displaystyle k\, \geq\, 3\)

I don't understand the question. could some one explain me this. I am messing up with "in each interval"? Thank you

Read it as "Show that one (or more) of the following is true
(A) If k\(\displaystyle \lt\)3 then f(x) is increasing
(B) If k\(\displaystyle \le\)3 then f(x) is increasing
..."
or, as in the case of multiple choice tests with only one answer, which of the above is the most inclusive statement.

 

[stapel]

EM 2.39 . .If \(\displaystyle f(x)\, =\, kx^3\, -\, 9x^2\, +\, 9x\, +\, 3\) is increasing in each interval, then:

. . . . . . . . . .(A) \(\displaystyle k\, <\, 3\) . . . (B) \(\displaystyle k\, \leq\, 3\) . . . (C) \(\displaystyle k\, >\, 3\) . . . (D) \(\displaystyle k\, \geq\, 3\)

I don't understand the question. could some one explain me this. I am messing up with "in each interval"? Thank you

If you take the first and second derivatives, set them equal to zero, and solve, you'll get the endpoints of intervals of sign. These intervals, along with their signs, will indicate "increasing", "decreasing", "concave up", and "concave down", or some selection of these, depending upon the function.

This particular question is saying that, once you've found these interval endpoints (in terms of k) for this particular function, figure out what restriction(s) you need to have on k so that, as a result, the signs on all of the intervals indicated by the first derivative will be positive, so that (at least between the endpoints, "in" each interval), the function f(x) is increasing.

So what did you get for the first derivative? When you set this equal to zero, what did you get? What can you do with this result? Please be complete. Thank you!

 

[Steven G]

EM 2.39 . .If \(\displaystyle f(x)\, =\, kx^3\, -\, 9x^2\, +\, 9x\, +\, 3\) is increasing in each interval, then:

. . . . . . . . . .(A) \(\displaystyle k\, <\, 3\) . . . (B) \(\displaystyle k\, \leq\, 3\) . . . (C) \(\displaystyle k\, >\, 3\) . . . (D) \(\displaystyle k\, \geq\, 3\)

I don't understand the question. could some one explain me this. I am messing up with "in each interval"? Thank you

We know that a general cubic \(\displaystyle f(x)\, =\, ax^3\, +\, bx^2\, +\, cx\, +\, d\) has no max or min when b^2-3ac<=0. In this case b^2-3ac = (-9)^2-3(k)(9)=81-27k<=0 or k>=3. But we know when the coefficient of the x^3 term is more than 0 and the cubic has no rel max or rel min then the function is strictly increasing. So where can we go from here?

 

[bhuvaneshnick]

[COLOR=#333333]Read it as "Show that one (or more) of the following is true[/COLOR]
[COLOR=#333333](A) If k[/COLOR][COLOR=#888888][/COLOR][COLOR=#333333][FONT=MathJax_Main]<[/FONT][/COLOR][COLOR=#333333]3 then f(x) is increasing[/COLOR]
[COLOR=#333333](B) If k[/COLOR][COLOR=#888888][/COLOR][COLOR=#333333][FONT=MathJax_Main]≤[/FONT][/COLOR][COLOR=#333333]3 then f(x) is increasing[/COLOR]
[COLOR=#333333]..."[/COLOR]
[COLOR=#333333]or, as in the case of multiple choice tests with only one answer, which of the above is the most inclusive statement.[/COLOR]

its increasing in all interval the how can i choose any one.See this is what i solved

differentiating first time and equating to 0
0=3kx² -18x+9
now do i want to find k,if so do i need to substitute value for x from the these four intervals.if doing so for every substitution i am getting different value of k.okay then with what interval i have to check whether its increasing?
what i am confusing with this is ,whether we have to consider the given interval for finding k or for to checking increasing by substituting in x in the equation
i dont have basic idea of how to solve.if there is no 'k' in the question i would have solved that with given interval.but this 'k' making me confuse

 

[Steven G]

[COLOR=#333333]Read it as "Show that one (or more) of the following is true[/COLOR]
[COLOR=#333333](A) If k[/COLOR][COLOR=#333333][FONT=MathJax_Main]<[/FONT][/COLOR][COLOR=#333333]3 then f(x) is increasing[/COLOR]
[COLOR=#333333](B) If k[/COLOR][COLOR=#333333][FONT=MathJax_Main]≤[/FONT][/COLOR][COLOR=#333333]3 then f(x) is increasing[/COLOR]
[COLOR=#333333]..."[/COLOR]
[COLOR=#333333]or, as in the case of multiple choice tests with only one answer, which of the above is the most inclusive statement.[/COLOR]

its increasing in all interval the how can i choose any one.See this is what i solved

differentiating first time and equating to 0
0=3kx² -18x+9
now do i want to find k,if so do i need to substitute value for x from the these four intervals.if doing so for every substitution i am getting different value of k.okay then with what interval i have to check whether its increasing?
what i am confusing with this is ,whether we have to consider the given interval for finding k or for to checking increasing by substituting in x in the equation
i dont have basic idea of how to solve.if there is no 'k' in the question i would have solved that with given interval.but this 'k' making me confuse

f(x) is increasing whenever f'(x)>0.
Ok, so you found f'(x). Very good.

Now f'(x) = 3kx² -18x+9 = 3(kx^2-6x+3). A cubic is strictly increasing if the coefficient of x^3 is positive and there is no rel max/min. DRAW cubics so you believe this 100%!!

The relative max/min occur when f'(x)=0. But we do not want this to happen! 3(kx^2-6x+3)=0 ->(kx^2-6x+3)=0

so x=[6 +/- sqrt(36-4(k)(3)]/2k. For x not to exist we want 36-4(k)(3)=36-12k=12(3-k) to be negative and this happens when 3-k<0 or k>3. Now what happens when k=3?

so x= [6+0]/(2k) = 3/k. You can't have a cubic with just a max or min but not both. If it matters, a quick inspection will show that f"(3/k) = 0 and when x=3/k we have a pt of inflection. So f is strictly increasing when k=3. So the final answer is k>=3

 

[bhuvaneshnick]

f'(x) = 3kx² -18x+9. So we want to find all values of K so f'(x)>0.

you mean that f'(x)>0 for -infinity <x<infinty ?

to find the value of x for min quadratic is x=-b/2a.i heard this from only you,thanks

Now f'(x) is a quadratic. Now the only why a quadratic could always be positive is if the coefficient of x^2, namely 3k, is positive. So we know k must be positive.

if f`(x) is cubic , you would have said that k should negative ? or you would have decided that based on from equation?

 

[Ishuda]

you mean that f'(x)>0 for -infinity <x<infinty ?

to find the value of x for min quadratic is x=-b/2a.i heard this from only you,thanks




if f`(x) is cubic , you would have said that k should negative ? or you would have decided that based on from equation?

Note that the k > 0 is important because ax² + bx + c has either a min OR max at x = -b/(2a). What determines whether it is a min or max is the sign of the second derivative which is 2a. Since a = 3k and k is positive, the value is a min at x=-b/(2a)

 

[Steven G]

Note that the k > 0 is important because ax² + bx + c has either a min OR max at x = -b/(2a). What determines whether it is a min or max is the sign of the second derivative which is 2a. Since a = 3k and k is positive, the value is a min at x=-b/(2a)

Thanks for explaining that better than me. It's all about team work to get through to these students.

 

[Steven G]

to find the value of x for min quadratic is x=-b/2a.i heard this from only you,thanks

This x =-b/(2a) value is the max or min of a quadratic. You definitely heard this before. But as a calculus student you can now understand the proof!

Let y= ax^2+bx+c. The max or min will be when the derivative = 0 (although you should check that it is not a pt of inf which actually is absurd for a quadratic)

Set y' = 2ax + b =0

Solving 2ax + b = 0 yields that x=-b/(2a)

 

[bhuvaneshnick]

thank you all

 

[Steven G]

you mean that f'(x)>0 for -infinity <x<infinty ? Yes

to find the value of x for min quadratic is x=-b/2a.i heard this from only you,thanks




if f`(x) is cubic , you would have said that k should negative ? or you would have decided that based on from equation? But f'(x) is quadratic

.

 

[stapel]

differentiating first time and equating to 0
0=3kx² -18x+9
now do i want to find k...?

I apologize for the confusion in this thread. To answer your question:

Yes, now that you've found the derivative of the original function and have set the results equal to zero, you want to find the values for k which ensure that the derivative is never negative. So you plug this quadratic into the Quadratic Formula:

. . . . .\(\displaystyle x\, =\, \dfrac{-(-18)\, \pm\, \sqrt{(-18)^2\, -\, 4(3k)(9)\,}}{2(3k)}\, =\, \dfrac{18\, \pm\, \sqrt{324\, -\, 108k\,}}{6k}\)

But, really, you only need the discriminant (the part inside the square root), because that's the part that determines the type of solution that the quadratic has: two different real numbers, one real number repeated, or complex values.

In order for the function to always be increasing "in" each interval, the derivative must be positive "in" each interval. For that to be the case, the derivative can only be zero at the interval endpoints, and the derivative must be positive everywhere else. If a quadratic has two real-number solutions, then you know (from experience) that the graph crosses the x-axis. In such a case, a positive quadratic (that is, an upward-opening parabola) must pass below the x-axis and take on negative y-values. This is not what you want in this case.

If the quadratic has complex-valued solutions, then you know (from experience) that the graph never touches the x-axis. Then the derivative, in this case, would be positive everywhere. But the exercise only specified that the derivative needs to be positive "in" the intervals; your solution needs to allow for the possibility that the derivative is zero at the interval endpoints.

The discriminant, the "324 - 108k" part of the solution expression, gives complex-valued solutions when it is negative, gives two different real-valued solutions when it is positive, and gives one (repeated) real-valued solution when it equals zero. In your case, what sort of solution must the quadratic equation have? So what sort of value must you have inside the square root?

Yes, 324 - 108k must equal zero in order to allow the derivative to equal zero or be less than zero in order to require the derivative always to be positive. So solve that inequality to find your solution.

 

[HallsofIvy]

This x =-b/(2a) value is the max or min of a quadratic. You definitely heard this before. But as a calculus student you can now understand the proof!

Let y= ax^2+bx+c. The max or min will be when the derivative = 0 (although you should check that it is not a pt of inf which actually is absurd for a quadratic)

Set y' = 2ax + b =0

Solving 2ax + b = 0 yields that x=-b/(2a)

You should be able to prove that before taking calculus: complete the square!

y= ax^2+ bx+ c= a(x^2+ (b/a)x)+ c= a(x^2+ (b/a)x+ b^2/(4a^2)- b^2/(4a^2))+ c

= a(x^2+ (b/a)x+ b^2/(4a^2))- b^2/4a+ c= a(x+ b/(2a))^2+ c- b^2/(4a).

That is -b^2/(4a) when x= -b/(2a) and, since a square is never negative, for any other x is, if a> 0, larger (so that x=-b/(2a) gives the function minimum) or, if a< 0, less (so that x= -b/(2a) gives the function maximum.