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
Help would be GREATLY appreciated !
The value of the function at x = -3 is
-27 -27/2+18+10 = -12.5
at x = -2 it is
-8 -12/2 +12+10 = 8
the function is continuous so must pass through zero on the way from -12.5 to +8
dy/dx = 3 x^2 -3x -6
at x = 0, dy/dx = -6
so slope of our line = +1/6
at x = 0, y = 10
so find equation of the line that passes through (0,10) with slope m = 1/6
dy/dx = 0 = 3 x^2 - 3x -6
x^2 - x - 6 = 0
(x-3)(x+2) = 0
so parallel to x axis at x = +3 and x = -2
find y at those points.
That is actually incorrect for the last part. The equation should turn out to be x^2-X-3. and then you should factor that to get x=-1 and x=2
a) To determine if f(x) has a root between x = -3 and x = -2, we can analyze the behavior of the function at these points. First, evaluate f(-3) and f(-2).
f(-3) = (-3)^3 - (3/2)(-3)^2 - 6(-3) + 10 = -46.5
f(-2) = (-2)^3 - (3/2)(-2)^2 - 6(-2) + 10 = 19
Since the function values change sign between f(-3) and f(-2) (negative to positive), by the Intermediate Value Theorem, we can conclude that there must be 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 find the derivative of f(x) and determine its slope at x = 0. Then, we can use the slope-intercept form (y = mx + b) to write the equation.
First, find the derivative of f(x):
f'(x) = 3x^2 - 3x - 6
Next, evaluate f'(0) to determine the slope at x = 0:
f'(0) = 3(0)^2 - 3(0) - 6 = -6
Now, we know that the slope of the line perpendicular to the graph of f at x = 0 is the negative reciprocal of -6, which is 1/6.
Using the slope-intercept form, the equation of the line perpendicular to the graph at x = 0 becomes:
y = (1/6)x + b
To find the value of b, substitute the coordinates of the point (0, f(0)) into the equation:
0 = (1/6)(0) + b
0 = b
Therefore, the equation of the line perpendicular to the graph of f at x = 0 is:
y = (1/6)x
c) To find the point on the graph of f where the tangent is parallel to the x-axis, we need to find the x-coordinate, which is the root of the derivative f'(x). Therefore, we need to solve the equation f'(x) = 0.
f'(x) = 3x^2 - 3x - 6
Set f'(x) equal to 0 and solve for x:
0 = 3x^2 - 3x - 6
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Factoring is not feasible in this case, so let's use the quadratic formula.
x = (-(-3) ± √((-3)^2 - 4(3)(-6))) / (2(3))
x = (3 ± √(9 + 72)) / 6
x = (3 ± √(81)) / 6
x = (3 ± 9) / 6
From these solutions, we get two possible values for x: x = (3 + 9) / 6 = 2 and x = (3 - 9) / 6 = -1
Now, substitute these x-values back into the original function f(x) to find their corresponding y-coordinates:
For x = 2:
f(2) = (2)^3 - (3/2)(2)^2 - 6(2) + 10
f(2) = 8 - 6 - 12 + 10
f(2) = -2
So, the point on the graph of f where the tangent is parallel to the x-axis is (2, -2).
For x = -1:
f(-1) = (-1)^3 - (3/2)(-1)^2 - 6(-1) + 10
f(-1) = -1 - 3/2 + 6 + 10
f(-1) = 5.5
Therefore, the point on the graph of f where the tangent is parallel to the x-axis is (-1, 5.5).