We are given a cubic function $f(x) = ax^3 + bx^2 + cx + d$. We need to find the values of the coefficients $a, b, c,$ and $d$, given that the function has a local maximum at $x = 1$ and the graph of the function is tangent to the x-axis at $x = 3$. In the second part of the problem, we need to study the variation and sketch the graph of the function $f(x) = x^3 - 6x^2 + 9x$.

1. Problem Description

We are given a cubic function $f(x) = ax^3 + bx^2 + cx + d$. We need to find the values of the coefficients $a, b, c,$ and $d$, given that the function has a local maximum at $x = 1$ and the graph of the function is tangent to the x-axis at $x = 3$. In the second part of the problem, we need to study the variation and sketch the graph of the function $f(x) = x^3 - 6x^2 + 9x$.

2. Solution Steps

Part 1: Find the coefficients $a, b, c, d$ of $f(x) = ax^3 + bx^2 + cx + d$ given that $f(x)$ has a local maximum at $x = 1$ and the graph is tangent to the x-axis at $x = 3$. We also know that at the tangent point, $y = 0$.

Since $f(x)$ is tangent to the x-axis at $x=3$, we have $f(3) = 0$.

$f(3) = a(3)^3 + b(3)^2 + c(3) + d = 27a + 9b + 3c + d = 0$ (1)

Since $f(x)$ has a local maximum at $x=1$, we have $f'(1) = 0$.

First, find the derivative of $f(x)$:

$f'(x) = 3ax^2 + 2bx + c$

Now, plug in $x=1$:

$f'(1) = 3a(1)^2 + 2b(1) + c = 3a + 2b + c = 0$ (2)

Since $f(x)$ is tangent to the x-axis at $x=3$, we also have $f'(3) = 0$.

$f'(3) = 3a(3)^2 + 2b(3) + c = 27a + 6b + c = 0$ (3)

Since the problem says $y=4$ for the local maxima, this is ambiguous as to where the 4 goes. However, since we know $x=1$, the point is $(1,4)$, so $f(1) = 4$.

$f(1) = a(1)^3 + b(1)^2 + c(1) + d = a + b + c + d = 4$ (4)

We have a system of 4 equations with 4 unknowns:

(1) $27a + 9b + 3c + d = 0$

(2) $3a + 2b + c = 0$

(3) $27a + 6b + c = 0$

(4) $a + b + c + d = 4$

Subtract (2) from (3):

$(27a + 6b + c) - (3a + 2b + c) = 0 - 0$

$24a + 4b = 0$

$6a + b = 0$

$b = -6a$

Substitute $b = -6a$ into (2):

$3a + 2(-6a) + c = 0$

$3a - 12a + c = 0$

$-9a + c = 0$

$c = 9a$

Substitute $b = -6a$ and $c = 9a$ into (1):

$27a + 9(-6a) + 3(9a) + d = 0$

$27a - 54a + 27a + d = 0$

$0 + d = 0$

$d = 0$

Substitute $b = -6a$, $c = 9a$, and $d = 0$ into (4):

$a + (-6a) + (9a) + 0 = 4$

$a - 6a + 9a = 4$

$4a = 4$

$a = 1$

Now find $b$ and $c$:

$b = -6a = -6(1) = -6$

$c = 9a = 9(1) = 9$

So, $a = 1, b = -6, c = 9, d = 0$.

Part 2: Study the variation and sketch the graph of $f(x) = x^3 - 6x^2 + 9x$.

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

To find critical points, set $f'(x) = 0$:

$3(x-1)(x-3) = 0$

$x = 1$ or $x = 3$

$f''(x) = 6x - 12$

$f''(1) = 6(1) - 12 = -6 < 0$, so $x = 1$ is a local maximum.

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

So, the local maximum is at $(1, 4)$.

$f''(3) = 6(3) - 12 = 18 - 12 = 6 > 0$, so $x = 3$ is a local minimum.

$f(3) = (3)^3 - 6(3)^2 + 9(3) = 27 - 54 + 27 = 0$

So, the local minimum is at $(3, 0)$.

3. Final Answer

$a = 1, b = -6, c = 9, d = 0$

The function is $f(x) = x^3 - 6x^2 + 9x$.

$f(x)$ has a local maximum at $(1, 4)$ and a local minimum at $(3, 0)$.