Finding value of expression with one stationary point

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Colin67

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Can somebody suggest how to proceed please, or have I gone off on the wrong track?

The question is:


1584305047390.png

So I know derivative at stationary value is zero, so I have differentiated the expression:

1584305168230.png

I have tried setting this expression to zero but get k in terms of x. Also I was thinking is the discriminant involved since it mentioned one stationary point, hence 1 root?

Many thanks in advance
 
I have checked and I cannot see my error. I have checked with Symbolab, it has the same answer.

1584307558680.png
 
OK, I guess it is correct. I would however factor out an (x+1) from the numerator--after all it is easily seen that x=-1 is a root of the numerator.
This will reduce the numerator from a cubic to a quadratic making the work a bit easier.

EDIT: If you prefer you can factor the numerator by grouping
 
Colin67 said:
I have checked and I cannot see my error. I have checked with Symbolab, it has the same answer.

View attachment 17192
Click to expand...
-2x^3+kx^2-2x^2-k

=-2x^2(x+1) +k(x^2-1)

=(x+1)(-2x^2 + kx - k)

Now the roots involving 'k' can be calculated from:

2x^2 - kx + k = 0

To have only one root (x = -1) for the original equation, the roots of the equation above must be "imaginary". So:

k^2 - 8k < 0 ...... and continue.....
 
Last edited by a moderator:
Subhotosh Khan said:
-2x^3+kx^2-2x^2-k

=-2x^2(x+1) +k(x^2-1)

=(x+1)(-2x^2 + kx - k)

Now the roots involving 'k' can be calculated from:

2x^2 - kx + k = 0

To have only one root (x = -1) for the original equation, the roots of the equation above must be "imaginary". So:

k^2 - 8k < 0 ...... and continue.....
Click to expand...
This is what happens when administrators tries to do the work of the workers--they mess up (To OP: this is all inside joking, but what follows is not)

Prof Khan, since (x+1) in the numerator gets absorbed by the (x+1)^4 in the denominator I have a problem with you saying that there is a stationary point at x=-1.
i would then conclude that to have exactly one root we need k^2 - 8k = 0. To OP, please solve this.

To Prof Khan--> Go directly to the corner for k^2 - 8k minutes, and I get to decide on what k equals.
 
I would write:

[MATH]f(x)=x(x+1)^{-2}(x-k)^{-1}[/MATH]
Hence:

[MATH]f'(x)=(x+1)^{-2}(x-k)^{-1}+x(-2(x+1)^{-3})(x-k)^{-1}+x(x+1)^{-2}(-(x-k)^{-2})=\frac{(x+1)(x-k)-2x(x-k)-x(x+1)}{(x+1)^3(x-k)^2}[/MATH]
[MATH]f'(x)=-\frac{2x^2-kx+k}{(x+1)^3(x-k)^2}[/MATH]
And then as advised, we equate the discriminant of the numerator to zero to ensure we have a repeated root:

[MATH](-k)^2-4(2)(k)=0[/MATH]
[MATH]k^2-8k=0[/MATH]
 
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