3) Dada la función f(x)=x^3-12x^2+36x-1 , determina: a. Los intervalos donde la función es creciente y decreciente.
b. Encuentra los números críticos.
c. Encuentra el intervalo donde la gráfica es cóncava hacia arriba y cóncava hacia abajo.
d. Encuentra los valores máximos y mínimos relativos (o locales).
e. Puntos máximos y mínimos relativos (o locales).

jml angle lkj & angle mj = lm

i have no clue what you just said. write a question in english?

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.