Pedagogy (thread 1)

[Steven G]

From this thread in "Calculus":

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

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. I agree with this.

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. Not true. Just consider when k<0
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. This is exactly what I said when I wrote We know that a general cubic [FONT=MathJax_Math]f[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]c[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math]d[/FONT][/FONT]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?

Here is another response I made as I thought that the student may not have known the formula for the x values for the max and min of any cubic (that has a max and min)

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%!! I feel that this is obvious

The relative max/min occur when f'(x)=0. You said this
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
What is not clear/correct in this response?

 

[stapel]

I understand that you enjoy engaging in pedagogical discussions, posting tutoring articles, and providing links to your tutoring site. However, it would be appreciated if you could kindly please restrict those activities to areas other than where students are trying to get help.

Thank you for your consideration.

 

[Steven G]

I understand that you enjoy engaging in pedagogical discussions, posting tutoring articles, and providing links to your tutoring site. However, it would be appreciated if you could kindly please restrict those activities to areas other than where students are trying to get help.

Thank you for your consideration.

Not a problem. But please do not say that I posts links to my website as this is simply not true. And for the record I did not plan on posting any more tutoring articles.