SOLUTION: Mediante GeoGebra obtener raíces, puntos máximos y mínimos, así como las gráficas de las siguientes funciones. p(x)=x^3 -7 x^2+15x-9 p(x)=x^4+ x^3 -4 x^2-4x p(x)=x^

Algebra ->  Linear-equations -> SOLUTION: Mediante GeoGebra obtener raíces, puntos máximos y mínimos, así como las gráficas de las siguientes funciones. p(x)=x^3 -7 x^2+15x-9 p(x)=x^4+ x^3 -4 x^2-4x p(x)=x^      Log On


   



Question 1176786: Mediante GeoGebra obtener raíces, puntos máximos y mínimos, así como las gráficas de las siguientes funciones.
p(x)=x^3 -7 x^2+15x-9
p(x)=x^4+ x^3 -4 x^2-4x
p(x)=x^5-10x^3 +9x

Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(53763) About Me  (Show Source):
You can put this solution on YOUR website!
.

Use  GeoGebra,  as assigned,  and complete this task  ON  YOUR  OWN.


Do not re-load  YOUR  JOB  to the shoulders of others.



Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!
1.
p%28x%29=x%5E3+-7x%5E2%2B15x-9.......factor completamente
p%28x%29=x%5E3+-x%5E2-6x%5E2%2B6x%2B9x-9
p%28x%29=%28x%5E3+-x%5E2%29-%286x%5E2-6x%29%2B%289x-9%29
p%28x%29=x%5E2%28x+-1%29-6x%28x-1%29%2B9%28x-1%29
p%28x%29+=+%28x%5E2+-+6x%2B9%29+%28x+-+1%29
p%28x%29+=+%28x+-+3%29%5E2+%28x+-+1%29
soluciones:
%28x+-+3%29%5E2+%28x+-+1%29=0
si %28x+-+3%29%5E2+=0+=>%28x+-+3%29+=0 =>x+=3
si %28x+-+1%29=0+=> x=1

Suppose that x=c is a critical point of f (x ) then:
If f'(x)>0 to the left of x=c and f' (x )<0 to the right of x=c then x=c is a local maximum.
If f'(x) < 0 to the left of x=c and f' (x )>0 to the right of x=c then x=c is a local minimum.
If f (x ) is the same sign on both sides of x=c then x=c is neither a local maximum nor a local minimum.
Suponga que x = c es un punto crítico de f (x) entonces:
Si f '(x)> 0 a la izquierda de x = c y f' (x) <0 a la derecha de x = c, entonces x = c es un máximo local.
Si f '(x) <0 a la izquierda de x = c y f' (x)> 0 a la derecha de x = c, entonces x = c es un mínimo local.
Si f (x) es el mismo signo en ambos lados de x = c, entonces x = c no es ni un máximo local ni un mínimo local.
entonces, encuentre la primera derivada de la función p (x)
p'%28x%29=3x%5E2+-14x%2B15
3x%5E2+-14x%2B15=0
%28x+-+3%29+%283+x+-+5%29+=+0
x-3=0 ->x=3
3x-5=0+-> x=5%2F3
ahora encuentra
p%28x%29=x%5E3+-7x%5E2%2B15x-9 if x=3
p%283%29=3%5E3+-7%2A3%5E2%2B15%2A3-9=0
(3,0)
and if x=5%2F3
p%285%2F3%29=%285%2F3%29%5E3+-7%2A%285%2F3%29%5E2%2B15%2A%285%2F3%29-9=32%2F27
(5%2F3,32%2F27)
entonces usted tiene
Maximum: (5%2F3,32%2F27)
Minimum:(3,0)

2.
p%28x%29=x%5E4%2B+x%5E3+-4+x%5E2-4x
p%28x%29+=+x+%28x%5E3+%2B+x%5E2+-+4+x+-+4%29
p%28x%29+=+x+%28%28x%5E3+%2B+x%5E2%29+-+%284+x%2B+4%29%29
p%28x%29+=+x+%28x%5E2%28x+%2B+1%29+-+4%28x%2B+1%29%29
p%28x%29+=+x+%28%28x%5E2+-+4%29%28x%2B+1%29%29
p%28x%29+=+x%28x+-+2%29%28x+%2B+2%29%28x+%2B+1%29
soluciones:
x=0
x=2
x=-2
x=-1
p'%28x%29=4x%5E3%2B+3x%5E2+-8x-4
4x%5E3%2B+3x%5E2+-8x-4=0
usando la calculadora obtendrás
x-1.61
x-0.47
x1.33
p%28-1.61%29=%28-1.61%29%5E4%2B+%28-1.61%29%5E3+-4+%2A%28-1.61%29%5E2-4%2A%28-1.61%29=-1.38->(-1.61,-1.38)
p%28-0.47%29=%28-0.47%29%5E4%2B+%28-0.47%29%5E3+-4+%2A%28-0.47%29%5E2-4%2A%28-0.47%29=0.94->(-0.47,0.94)
p%281.33%29=%281.33%29%5E4%2B+%281.33%29%5E3+-4+%2A%281.33%29%5E2-4%2A%281.33%29=-6.91->(1.33,-6.91)
Maximum:(-0.47,0.94)
Minimum:(1.33,-6.91)

3.
p%28x%29=x%5E5-10x%5E3+%2B9x
p%28x%29=x%28x%5E4-10x%5E2+%2B9%29
p%28x%29=x%28x%5E4-x%5E2-9x%5E2+%2B9%29
p%28x%29=x%28%28x%5E4-x%5E2%29-%289x%5E2+-9%29%29
p%28x%29=x%28x%5E2%28x%5E2-1%29-9%28x%5E2+-1%29%29
p%28x%29=x%28%28x%5E2-9%29%28x%5E2+-1%29%29
p%28x%29=x%28%28x%5E2-3%5E2%29%28x%5E2+-1%29%29
p%28x%29=x%28x-3%29%28x%2B3%29%28x+-1%29%28x%2B1%29
=> soluciones:
x=0
x=3
x=-3
x=1
x=-1
p'%28x%29=5x%5E4-30x%5E2+%2B9
5x%5E4-30x%5E2+%2B9=0
5x%5E4-30x%5E2+%2B9=0
usando la calculadora obtendrás
x+=+sqrt%283+-+6%2Fsqrt%285%29%29x=0.56
x+=+-sqrt%283+-+6%2Fsqrt%285%29%29x=-0.56
x+=+sqrt%283%2B6%2Fsqrt%285%29%29x=2.38
x+=+-sqrt%283+%2B+6%2Fsqrt%285%29%29x=-2.38

p%280.56%29=%280.56%29%5E5-10%2A%280.56%29%5E3+%2B9%280.56%29=3.34 ->(0.56,3.34)
p%28-0.56%29=%28-0.56%29%5E5-10%2A%28-0.56%29%5E3+%2B9%28-0.56%29=-3.34 ->(-0.56,-3.34)
p%282.38%29=%282.38%29%5E5-10%2A%282.38%29%5E3+%2B9%282.38%29=-37.03 ->(2.38,-37.03)
p%28-2.38%29=%28-2.38%29%5E5-10%2A%28-2.38%29%5E3+%2B9%28-2.38%29=37.03 ->(-2.38,37.03)

Maximum:(0.56,3.34) y (-0.56,-3.34)
Minimum:(2.38,-37.03) y (2.38,-37.03)