# Thread: Relative Maximums and Minimums

1. ## Relative Maximums and Minimums

let $\displaystyle P(x)= x^4+ax^3+bx^2+cx+d$
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

2. Originally Posted by Godfather
let $\displaystyle P(x)= x^4+ax^3+bx^2+cx+d$
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.

3. $\displaystyle P(x)= x^4+ax^3+bx^2+cx+d$
Extremum
$\displaystyle \frac{dP}{dx}= 4x^3+3ax^2+2bx+c$
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).
$\displaystyle \frac{dP}{dx}= \alpha x (x-q)(x+q) = \alpha (x^3 - q^2 x)$
Deduce that $\displaystyle \alpha = 4 , a = c = 0 \text { and } 2b = -\alpha q^2 = -4q^2 \implies b = -2q^2$
$\displaystyle P(x)= x^4-2q^2 x^2+d$
$\displaystyle P(0)= 0^4-2q^2 0^2+d = d = 1$
$\displaystyle P(q)= q^4-2q^4+1 = -3$
$\displaystyle P(q)= -q^4 = -4$
$\displaystyle P(q)= q^2 = 2$
-2 doesn't work here that why we omit the ± sign.
$\displaystyle P(q)= q = \pm \sqrt{2}$

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

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