differentiating question: let f(x)=x^3-3x^2+kx+8 where k is a constant.

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ontopoftheworld75

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hi this is a hsc question from a past paper

let f(x)=x^3-3x^2+kx+8 where k is a constant.Find the values of k for which f (x) is an increasing function.

so i guess i know that f'(x)≠0
and f'(x)>0
and f"(x)≠0
but is f"(x)>0 or just not equal to zero?

so after differentiating i got f'(x)=3x^2-6x+k
hence 3x^2-6x+k>0

then f"(x)=6x-6
let f"(x)=0 (PLEASE TELL ME IF THIS IS CORRECT AND THE RIGHT WAY TO SET MY WORKING OUT)
hence 6x=6
therefore x=1

sub x=1 into f'(x)
f'(1)=3-6+k
k=3

so the answer is k>3 but i got 3 but i don't know how to go about approaching these questions and what the formal working out should look like?
it would be great if someone could post full working out so i know how to approach these questions and the steps involved i guess,
Thanks so much for reading all this your help if greatly appreciated!
regards
 
ontopoftheworld75 said:
Let f(x) = x^3 - 3x^2 + kx + 8 where k is a constant. Find the values of k for which f(x) is an increasing function.

so i guess i know that f'(x)≠0
and f'(x)>0
and f"(x)≠0
but is f"(x)>0 or just not equal to zero?
Click to expand...
Why would the second derivative need to be positive? For instance, the log functionn is concave down, yet is always increasing.

ontopoftheworld75 said:
so after differentiating i got f'(x) = 3x^2 - 6x + k
hence 3x^2 - 6x + k > 0
Click to expand...
Yes. It is this quadratic inequality which must be solved.

ontopoftheworld75 said:
then f"(x)=6x-6
let f"(x)=0 (PLEASE TELL ME IF THIS IS CORRECT AND THE RIGHT WAY TO SET MY WORKING OUT)
hence 6x=6
therefore x=1
Click to expand...
This solves for the (potential) inflection point. But this tells you nothing about the value of "k".

ontopoftheworld75 said:
sub x=1 into f'(x)
f'(1)=3-6+k
k=3
Click to expand...
How? This works only if f'(1) = 0, but you are not given this information. In fact, you have assumed that this equality is false! :shock:

Instead, let's work with the givens (the function and the requirement that it be an increasing function) and the algebra.

You have:

. . . . .\(\displaystyle f'(x)\, =\, 3x^2\, -\, 6x\, +\, k\)

. . . . .\(\displaystyle 3x^2\, -\, 6x\, +\, k\, >\, 0\)

The derivative is a positive quadratic, so you know that the graph is an upward-opening parabola. Thus, it can be negative only between its x-intercepts. Using this algebraic fact, find these zeroes:

. . . . .\(\displaystyle 3x^2\, -\, 6x\, +\, k\, =\, 0\)

Since we don't have a value for "k" (because it's what we need to solve for), we'll have to use the Quadratic Formula:

. . . . .[math]x\, =\, \dfrac{-(-6)\, \pm\, \sqrt{(-6)^2\, -\, 4(3)(k)\,}}{2(3)}\, -\, \dfrac{6\, \pm\, \sqrt{36\, -\, 12k\,}}{6}[/math]
. . . . .[math]\, =\, \dfrac{6\, \pm\, 2\, \sqrt{9\, -\, 3k\,}}{6}\, =\, \dfrac{3\, \pm\, \sqrt{9\, -\, 3k\,}}{3}[/math]

The relevant point is that we want there to be no real zeroes (that is, no real-valued solutions to the above) so that there are no graphed x-intercepts. This is because we want the derivative (that quadratic that we've been working with) always to be positive; in particular, we want it never to equal zero.

From working in the past with graphing and the Quadratic Formula, you know that the only way for this quadratic derivative never to equal zero is for the square root to have a negative inside the square root. In other words, you need:

. . . . .\(\displaystyle 9\, -\, 3k\, <\, 0\)

With this restriction, your original function should fulfill the requirement. If you're not sure, try plugging various values in for "k", and see what kinds of graphs you get back. ;)
 
"
. . . . .[imath]\displaystyle x\, =\, \dfrac{-(-6)\, \pm\, \sqrt{\strut (-6)^2\, -\, 4(3)(k)\,}}{2(3)}\, -\, \dfrac{6\, \pm\, \sqrt{\strut 36\, -\, 12k\,}}{6}[/imath]

. . . . .[imath]\displaystyle \, =\, \dfrac{6\, \pm\, 2\, \sqrt{\strut 9\, -\, 3k\,}}{6}\, =\, \dfrac{3\, \pm\, \sqrt{\strut 9\, -\, 3k\,}}{3}[/imath]
"

and I don't get the negative inside of the root. I thought you can't have a negative in a root.
 
Last edited:
AlonzoN said:
I don't get the negative inside of the root
Click to expand...
Hello Alonzo. Please see post #2 again. A recent change to the forum software has broken the math formatting in many threads. We're repairing these formatting errors as they come to our attention. We regret the inconvenience.

?
 
I would approach it a little differently.
The graph of f(x) will be increasing if f ' (x) > 0 as you said.

So \(\displaystyle 3x^2-6x +k>0\) as you also said.

\(\displaystyle y=3x^2 -6x+k\) is a parabola with a minimum turning point (since 3>0). So you want this TP to lie above the x-axis.

The TP occurs at \(\displaystyle x=\frac{-b}{2a} = \frac{6}{6}=1\) or this can be found by letting f ''(x)=0 as you did.
therefore the TP is (1, -3+k) .

This TP is above the x-axis when k>3.
 
Harry_the_cat said:
I would approach it a little differently.
The graph of f(x) will be increasing if f ' (x) > 0 as you said.

So \(\displaystyle 3x^2-6x +k>0\) as you also said.

\(\displaystyle y=3x^2 -6x+k\) is a parabola with a minimum turning point (since 3>0). So you want this TP to lie above the x-axis.

The TP occurs at \(\displaystyle x=\frac{-b}{2a} = \frac{6}{6}=1\) or this can be found by letting f ''(x)=0 as you did.
therefore the TP is (1, -3+k) .

This TP is above the x-axis when k>3.
Click to expand...
Thank you HEAPS!!!
 
AlonzoN said:
Also, I thought that there can't be a negative in a square root.
Click to expand...
Yes and no. You can have a 0 under the square root sign, but it will NOT be a real number. It will be what is called a complex number. If a quadratic has a complex root/zero/x-intersect, then this quadratic function does not cross the x-axis. That is what we want! This only happens if what is under the square root sign is negative.
 
oh. Hey, I have another post, can you help me out? Harry the cat was helping me, but I think he left.
 
Jomo said:
Yes and no. You can have a 0 under the square root sign, but it will NOT be a real number. It will be what is called a complex number. If a quadratic has a complex root/zero/x-intersect, then this quadratic function does not cross the x-axis. That is what we want! This only happens if what is under the square root sign is negative.
Click to expand...
oh. Hey, I have another post, can you help me out? Harry the cat was helping me, but I think he left.
 
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