Relative Maximums and Minimums

**Godfather** · November 29th 2008, 02:47 PM

let $P(x)= x^4+ax^3+bx^2+cx+d$
The graph of y=P(x) is symmetric with respect to the Y Axis, the relative maximum at (0,1) and has an absolute minimum at (q,-3).

(A). Determine the values of a,b,c and d and using these values write an expression for P(x)
(B). find all possible values of q

**skeeter** · November 29th 2008, 02:57 PM

Originally Posted by Godfather

let $P(x)= x^4+ax^3+bx^2+cx+d$
The graph of y=P(x) is symmetric with respect to the Y Axis, the relative maximum at (0,1) and has an absolute minimum at (q,-3).

(A). Determine the values of a,b,c and d and using these values write an expression for P(x)
(B). find all possible values of q

P(x) is symmetrical to the y-axis ... that should tell you what the coefficients "a" and "c" are.

max at (0,1) tells you two things ... the value of "d" and that P'(x) = 0

abs min at (q, -3) is also at (-q, -3) ... graph is symmetrical.
also, P'(x) = 0 at x = q and x = -q.

work on it.

**vincisonfire** · November 29th 2008, 05:28 PM

$ P(x)= x^4+ax^3+bx^2+cx+d $
Extremum
$ \frac{dP}{dx}= 4x^3+3ax^2+2bx+c $
Relative maximum at (0,1) and has an absolute minimum at (q,-3).
Since the function is symmetric there is also an absolute minimum at (-q,-3).
$ \frac{dP}{dx}= \alpha x (x-q)(x+q) = \alpha (x^3 - q^2 x) $
Deduce that $\alpha = 4 , a = c = 0 \text { and } 2b = -\alpha q^2 = -4q^2 \implies b = -2q^2$
$ P(x)= x^4-2q^2 x^2+d $
$ P(0)= 0^4-2q^2 0^2+d = d = 1 $
$ P(q)= q^4-2q^4+1 = -3 $
$ P(q)= -q^4 = -4 $
$ P(q)= q^2 = 2 $
-2 doesn't work here that why we omit the ± sign.
$ P(q)= q = \pm \sqrt{2} $

**Godfather** · November 30th 2008, 04:40 PM

how did u deduce that formula after the first derivitive? where did the alpha come in? how did u do that?

**vincisonfire** · November 30th 2008, 04:46 PM

$ \frac{dP}{dx}= \alpha x (x-q)(x+q) = \alpha (x^3 - q^2 x) $
When x is equal to $q$ or $-q$ , it is given that the first derivative is equal to 0 since there are local minima there.
Therefore, the polynomial will have the form $\frac{dP}{dx}= f(x)(x-q)(x+q)$ so that it equals 0 when x is equal to $q$ or $-q$ .
Then you get that $\frac{dP}{dx}= g(x)(x)(x-q)(x+q)$ since there is a local max at x=0.
You know that your polynomial is order three. Therefore, g(x) = constant.

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