Calculus Help?

Question

given f(x) = x^3 - 6x^2 + 9x and g(x) = 4
1) find the coordinates of the points common to the graphs of f and g
2) find all the zeros of f
3) If the domain of f is limited to the close interval [0,2], what is the range of f?

Pleas Help, thx in advance

Answer 1

to answer (1) find the points where f(x) = 4 since for any x, g(x) = 4

4 = x^3 - 6x^2 + 9x
x(x^2 - 6x + 9) = 4
x(x-3)(x-3) = 4
x = 4 and x = 1

therefore the points common to f and g are (4,4) and (1,4)
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f(x) = 0 ie. find roots
x^3 - 6x^2 + 9x = 0
x(x-3)(x-3) = 0
x=0, x =3
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if the domain of f is the closed interval [0,2] then
the range of f is the minimum and maximum value in that range. To find the minimum and maximum take the derivative

f'(x) = 3x^2 - 12x + 9
there is a min or max when f'(x) = 0
3x^2 - 12x + 9 =0
(3x - 3)(x - 3) =0
x = 1 and x = 3
since x=3 is outside the domain it is rejected

f(1) = 1^3 - 6(1)^2 + 9(1)
f(1) = 1-6+9 = 4

also check both endpoints of the domain
f(0) = 0 and f(2) = 2^3 - 6(2)^2 + 9(2) = 2

therefore the range of f on the domain [0,2] is 0,4

Answer 2

1) To find the common points, we need only find the values where f(x) has the value 4 = g(x). We want to know where x^3 - 6x^2 + 9x = 4; this is the same as solving the equation
x^3 - 6x^2 + 9x -4 = 0. We find that 4 is a solution of this equation, and it is the only real solution. Therefore, the point (4,4) is the only point in common.

2) To find the zeros of f, we notice that f(x) = x(x - 3)^2. Therefore x = 0 and x = 3 are the only zeros of f.

3) We look at the derivative of f and find that f'(x) = 3x^2 - 6x + 9. Since the zeros are not real, this derivative never changes sign. In fact, f'(x) > 0 for all x. Therefore the function is always increasing, and in particular, it is strictly increasing on [0,2]. Thus, the range is [f(0),f(2)] = [0,2].

Answer 3

The graphs intersect when
x^3 - 6x^2 + 9x - 4 = 0
= x^3 - 4x^2 - 2x^2 + 9x - 4
= x^2 * (x-4) - (2x^2 - 9x + 4)
= x^2 (x-4) - (2x - 1)(x-4)
= (x^2 - 2x + 1)(x-4)
= (x-1)^2 (x-4) = 0
x = 1,4

The corresponding y value will be 4 in both cases, since g = 4
Points in common = (1,4), (4,4)

Answer 4

f(x)=x^3-6x^2+9x=x·(x^2-6x+9)=x·(x-3)^2
f'(x)=(3x^2-12x+9)=3·(x^2-4x+3)=3·(x-1)·(x-3)


f↑ from (-inf, -inf) to (1, 4) crossing (0,0)
f↓ from (1,4) to (3,0) and now
f↑ from (3,0) to (+inf, +inf)

1) f(x)=g(x) <--> f(x)=4 (two points)

x^3-6x^2+9x-4=0 ; (x-1)^2 ·(x-4)=0 -->

A(1,4) and B(4,4)

2) The zeros of f are x=0 and x= 3 --> Points C(0,0), D(3,0)

3) If 0 ≤x≤2 --> minf[0,2] ≤f(x)≤maxf[0,2] -->
0≤f(x)≤4 --> Range is [0, 4]

Saludos.

Answer 5

1) {1, 4}
2) {0, 3}
3) [0, 4]