Given the function defined as f(x)=x^3-(3/2)x^2-6x+10

a) Explain why f(x) must have a root between x=-3 and x=-2

b) Write an equation of the line perpendicular to the graph of f at x=0

c) Find the x and y coordinates of the point on the graph of f where the line tangent to the graph is parallel to the x-axis

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a) To determine whether f(x) has a root between x=-3 and x=-2, we can analyze the behavior of the function at the endpoints of this interval.

1. Evaluate f(-3):
f(-3) = (-3)^3 - (3/2)(-3)^2 - 6(-3) + 10
= -27 - (3/2)(9) + 18 + 10
= -27 - 13.5 + 18 + 10
= -12.5

2. Evaluate f(-2):
f(-2) = (-2)^3 - (3/2)(-2)^2 - 6(-2) + 10
= -8 - (3/2)(4) + 12 + 10
= -8 - 6 + 12 + 10
= 8

Since f(-3) is negative (-12.5) and f(-2) is positive (8), by the Intermediate Value Theorem of Calculus, f(x) must have at least one root between x = -3 and x = -2.

b) To find the equation of the line perpendicular to the graph of f at x = 0, we need to determine the slope of the tangent line to f(x) at x = 0 and then find the negative reciprocal of that slope.

1. Find the derivative of f(x):
f'(x) = 3x^2 - 3x - 6

2. Evaluate f'(0):
f'(0) = 3(0)^2 - 3(0) - 6
= -6

The slope of the tangent line to f(x) at x = 0 is -6.

3. The negative reciprocal of -6 is 1/6.

Therefore, the slope of the line perpendicular to the graph of f at x = 0 is 1/6.

c) To find the point on the graph of f where the tangent line is parallel to the x-axis, we need to find the x-coordinate of that point and then evaluate f(x) at that x-coordinate to obtain the y-coordinate.

1. To find the x-coordinate, equate the derivative of f(x) to zero and solve for x:
f'(x) = 3x^2 - 3x - 6
0 = 3x^2 - 3x - 6

Factoring the equation, we get:
0 = (3x + 2)(x - 2)

Solving for x, we have:
3x + 2 = 0 --> x = -2/3
x - 2 = 0 --> x = 2

Therefore, the x-coordinates of the points where the tangent line is parallel to the x-axis are x = -2/3 and x = 2.

2. Evaluate f(x) at those x-coordinates to obtain the y-coordinates:
f(-2/3) = (-2/3)^3 - (3/2)(-2/3)^2 - 6(-2/3) + 10
= -8/27 + 4/9 + 4 + 10
= -8/27 + 12/27 + 108/27 + 270/27
= 382/27

f(2) = (2)^3 - (3/2)(2)^2 - 6(2) + 10
= 8 - 6 + 4 - 12 + 10
= 4

Therefore, the coordinates of the points on the graph of f where the tangent line is parallel to the x-axis are (-2/3, 382/27) and (2, 4).

a) To determine why f(x) must have a root between x = -3 and x = -2, we need to use the Intermediate Value Theorem. According to the theorem, if a function is continuous over a closed interval [a, b], and it takes on values of opposite sign at the endpoints, then it must have at least one root in that interval.

In this case, we evaluate f(-3) and f(-2) to check if they have opposite signs. Let's calculate:

f(-3) = (-3)^3 - (3/2)(-3)^2 - 6(-3) + 10 = -27 - 13.5 + 18 + 10 = -12.5
f(-2) = (-2)^3 - (3/2)(-2)^2 - 6(-2) + 10 = -8 + 6 - 12 + 10 = -4

Since f(-3) = -12.5 is negative, and f(-2) = -4 is positive, we can conclude that f(x) changes sign between x = -3 and x = -2. Therefore, f(x) must have at least one root in this interval.

b) To find a line perpendicular to the graph of f at x = 0, we first need to determine the slope of the tangent line to the graph of f at x = 0. The slope of the tangent line is equivalent to the derivative of the function evaluated at x = 0.

Let's find the derivative of f(x):

f'(x) = 3x^2 - 3x - 6

Now, let's evaluate f'(0):

f'(0) = 3(0)^2 - 3(0) - 6 = -6

The slope of the tangent line to the graph of f at x = 0 is -6. Since the line perpendicular to this tangent line will have a slope that is the negative reciprocal of -6, we can find it by taking the negative reciprocal:

slope of the perpendicular line = -1/(-6) = 1/6

Now, we have the slope of the line perpendicular to the graph of f at x = 0. To find the equation of this line, we need a point it passes through. Since we know it passes through x = 0, we can use this as a point on the line.

Using the point-slope form of a line, the equation of the line perpendicular to the graph of f at x = 0 is:

y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point on the line.

Given (x1, y1) = (0, f(0)), substituting in the values:

y - f(0) = (1/6)(x - 0)
y - f(0) = (1/6)x
y = (1/6)x + f(0)

Therefore, the equation of the line perpendicular to the graph of f at x = 0 is y = (1/6)x + f(0).

c) To find the point on the graph of f where the line tangent to the graph is parallel to the x-axis, we need to find where the derivative of f(x) equals zero.

Let's find the derivative of f(x) again:

f'(x) = 3x^2 - 3x - 6

To find where the derivative equals zero, we set f'(x) = 0 and solve for x:

3x^2 - 3x - 6 = 0

Now we can solve this quadratic equation. Factoring or using the quadratic formula allows us to find the values of x:

3x^2 - 3x - 6 = 0
x^2 - x - 2 = 0 (dividing both sides by 3)

(x - 2)(x + 1) = 0 (factoring)

Setting each factor equal to zero:

x - 2 = 0 or x + 1 = 0

Solving for x:

x = 2 or x = -1

To find the corresponding y-coordinates, we substitute these x-values back into the original function f(x):

For x = 2:
f(2) = (2)^3 - (3/2)(2)^2 - 6(2) + 10 = 8 - 6 - 12 + 10 = 0

For x = -1:
f(-1) = (-1)^3 - (3/2)(-1)^2 - 6(-1) + 10 = -1 - (3/2) + 6 + 10 = 12.5

Hence, the coordinates where the tangent line is parallel to the x-axis are (2, 0) and (-1, 12.5).