Math LK Q1 – Function curves and locus curves?
Hello, I have a question about task 7.
I did find out that the extreme point is at x=((2k)^1/2)/2 (k is not 0 or 1/3) and the inflection point is at x=0 vx=((2/3)^1/2)/2. The locus of the minimum is y=-1/2*x^2 (0<k<1/3).
I have a question about c. Why doesn't the ratio depend on k? Is it because the locus is divided by a fixed inflection point?
Thanks!
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As a turning point, I find something different from you, with k below a root
fk”'(x) = 12×2 – 2k
x2 = 2/12 * k
x = wurz( 1/6 * k )
.
This cuts k (at xE also below root ) away
The location curve of the minimum is unfortunately not correct. This can be easily checked by using values. for k=4 is a low point x=sqrt(2). If this point is inserted into the original function f_k, the same must come out as when inserted into the local line.
f_k(sqrt(2)) ≠ y(sqrt(2))
-4 ≠-1
Determination of the local line:
First, the extreme points of the function f_k are determined, i.e. the zero points of the first derivative. This results in x1=0 and
Actually, it should be repeated with x3. As we potentize the negative root expression with straight exponents, for x3 exactly the same result comes out as for x2.
This function f(x2(k)) dependent on k can now be drawn badly into our x-y coordinate system. Therefore we must replace k with an expression with x:
We re-form our equation for the extreme x2 and x3 after k and get
In the dependent local line (we can call it g(k))
the k is now replaced by the above expression and we receive the local line as a function of x.
Part c:
The turning points (2nd part zero) are dependent on k as are the extreme points. If one divides extreme places by turning point or otherwise, the k is shortened and a constant remains. This means in the context: if we know the extreme values for all k, we can easily determine the turning points (and otherwise) without having to calculate the procedure with the derivation.
Thank you.