2 question on calculus?

Question

this is a hard problem not even my friend can figure it out

Let P(x) = x^4 +ax^3+bx^2=cx+d. The graph of y=p(x) is symmettric with respect to the y-axis, has a relative maximum at (0,1) and has an absolute maximum at (q,-3)

a) determine the values of a, b, c, and d, and using these values write an expression for p(x) (determine the values?)

b) Find all possible values of q. ( well how do i determine the values?)

and very quick question on particle movement

x=(2t-1)(t-1)^2 the particle moves on the x-axis

what interval is it moving to the left(does left mean pos to neg?)

and what time during that interval is particle moving most rapidly(speed is maximum?)

Answer 1

1) P(x) = x^4 +ax^3+bx^2+cx+d

firstly lets use the detail (has a relative maximum at (0,1)) to fidure out an equation

P(0)=1

this means d=1

so with this informatio P(x) = x^4 +ax^3+bx^2+cx+1

the detail also implies that P'(x)=0 when x=0

P(x) = x^4 +ax^3+bx^2+cx+d

P'(x)= 4x^3+3ax^2+2bx+c

P'(0)=c

so c=0

with this info P(x)=x^4 +ax^3+bx^2+1

now this detail (graph of y=p(x) is symmettric with respect to the y-axis) is a little bit tricky.

as answer 1 says P(x)=P(-x)

so x^4 +ax^3+bx^2+1=x^4 -ax^3+bx^2+1

this equals 2ax^3=0

now this means than "a" must be zero for the graph to be symmetric with respect to the y-axis

P(x)=x^4+bx^2+1

P'(x)=4x^3 +2bx

now there is a big MISTAKE IN YOU QUESTION. THE MISTAKE IS THAT A QUARTIC FUNCTION WITH A POSITVE LEADING COEFFICIENT CANNOT HAVE AN ABSOLUTE MAXIMUM BECAUSE IT EXTENDS FROM QUADRANT 2 TO QUADRANT 4.HOWEVER IT CAN HAVE A RELATIVE MAXIMUM OR AN ABSOLUTE MINIMUM.

P'(x)=4x^3 +2bx

P'(x)=2x(2x^2+b)

since at x=q there is an absolute maximum (which is incorrect, it should be absolute minimum)  P'(q)=0



P'(q)=2q(2q^2+b)

0= 2q(2q^2+b)

now we already know that there is a relative maximum at (0,1)

so our concern is (2q^2+b) which means b must be negative or else there wont be any more maximums or minimums. since the graph is symmetric about y-axis there will be 2 absolute minimums

2q^2=-b

q^2=-b/2

with this detail P(x)=x^4+bx^2+1

using (q,-3) -3=q^4+qb^2+1

q^4+qb^2+1=-3

q^4+qb^2=-4

since q^4+qb^2=-4  using q^2=-b/2

substitute

(-b/2)^2+b(-b/2)=-4

(b^2/4)-(b^2/2)=-4

(-b^2)/4=-4

(b^2)/4=4

b^2=16

b=-4 and 4

since b=4 does not work in our interest this is rejected

so b=-4

therefore  P(x)=x^4-4x^2+1

now part b) P(q)=-3

-3=q^4-4q^2+1

q^4-4q^2+4=0

((q^2)-2)^2=0

q^2=2

q=+/-sqrt(2)

x=(2t-1)(t-1)^2

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

x=2t^3 -4t^2+2t -t^2+2t-1

x=2t^3 -5t^2+4t-1

dx/dt=6t^2 -10t+4

particle moves on the left when dx/dt<0

dx/dt=6t^2 -6t-4t+4

dx/dt=6t(t-1)-4(t-1)

dx/dt=0

0=(6t-4)(t-1)

t=1 and t=2/3

so dx/dt<0 when 2/3<x<1

and the speed is maximum when the second derivative of x is equal to zero

Answer 2

if a function is symmetric about y axis then f(x) = f(-x). that'll give one equation. absolute max means a) 1st derivative zero at x=q and 2nd derivative less than 0 at x=q. of course f(q) = 3.

when t < 1/2 it moves to teh left

speed = dx/dt. just find the maxima of the function.

PS: you have dumb friends