this question concerns the function
f(x)=-2x3+3x2+12x+10
(a)find the stationary points of this function
using the strategy to apply the first derivative test,classify the left hand stationary point found in part (a)
find the y-coordinate of each of the stationary points on the graph of the function f(x), and also evaluate f(0)
To find the stationary points of a function, we first need to find the points where the first derivative is equal to zero. So let's compute the derivative of the given function:
f(x) = -2x^3 + 3x^2 + 12x + 10
f'(x) = d/dx(-2x^3) + d/dx(3x^2) + d/dx(12x) + d/dx(10)
= -6x^2 + 6x + 12
Now let's set f'(x) = 0 and solve for x:
-6x^2 + 6x + 12 = 0
=> x^2 - x - 2 = 0
Now we factor the quadratic equation:
(x - 2)(x + 1) = 0
So, we get two possible stationary points, x1 = 2 and x2 = -1.
Now to classify the left-hand stationary point using the first derivative test, we need to check the sign of the derivative around x1 = -1:
f'(-1.1) = 6.61 - 6.6 + 12 > 0 (It is positive)
f'(-0.9) = 5.61 - 5.4 + 12 > 0 (It is positive)
Since f'(-1.1) > 0 and f'(-0.9) > 0, we can conclude that the left-hand stationary point, x1 = -1, is a local minimum.
Now, we need the y-coordinate for both of the stationary points:
For x1 = -1:
f(-1) = -2(-1)^3 + 3(-1)^2 + 12(-1) + 10 = -2 + 3 - 12 + 10 = -1
For x2 = 2:
f(2) = -2(2)^3 + 3(2)^2 + 12(2) + 10 = -16 + 12 + 24 + 10 = 30
The stationary points on the graph of the function are (-1, -1) and (2, 30).
Now we need to evaluate f(0):
f(0) = -2(0)^3 + 3(0)^2 + 12(0) + 10 = 0 + 0 + 0 + 10 = 10
So, f(0) = 10.
To find the stationary points of a function, we need to first find the critical points. Critical points occur when the derivative of the function is equal to zero or undefined.
Step 1: Find the first derivative of the function f(x) with respect to x.
f'(x) = -6x^2 + 6x + 12
Step 2: Set the first derivative equal to zero and solve for x.
-6x^2 + 6x + 12 = 0
You can solve this quadratic equation using factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula.
Step 3: Apply the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a.
With a = -6, b = 6, and c = 12, the equation becomes:
x = (-6 ± √(6^2 - 4(-6)(12))) / (2(-6))
Simplifying further:
x = (-6 ± √(36 + 288)) / (-12)
x = (-6 ± √324) / -12
x = (-6 ± 18) / -12
This gives us two possible solutions for x:
x1 = (-6 + 18) / -12 = -12/12 = -1
x2 = (-6 - 18) / -12 = -24/12 = -2
Step 4: Evaluate the second derivative at each critical point to determine if it's a maximum, minimum, or neither.
To classify the left-hand stationary point (x = -2), we need to find the value of the second derivative at that point.
Take the second derivative of the function:
f''(x) = -12x + 6
Now substitute x = -2 into the second derivative:
f''(-2) = -12(-2) + 6
f''(-2) = 24 + 6
f''(-2) = 30
Since the second derivative at x = -2 is positive (30 > 0), it means the function is concave up at that point. Therefore, it is a local minimum.
Step 5: Find the y-coordinate of each stationary point on the graph.
To find the y-coordinate, substitute each critical point into the original function f(x).
For x = -1:
f(-1) = -2(-1)^3 + 3(-1)^2 + 12(-1) + 10
f(-1) = -2 + 3 - 12 + 10
f(-1) = -1
So, the left-hand stationary point is (-1, -1).
For x = -2:
f(-2) = -2(-2)^3 + 3(-2)^2 + 12(-2) + 10
f(-2) = -2(-8) + 3(4) - 24 + 10
f(-2) = 16 + 12 - 24 + 10
f(-2) = 14
So, the right-hand stationary point is (-2, 14).
Step 6: Evaluate f(0).
To find f(0), substitute x = 0 into the original function f(x):
f(0) = -2(0)^3 + 3(0)^2 + 12(0) + 10
f(0) = 0 + 0 + 0 + 10
f(0) = 10
So, f(0) = 10.
Therefore, the stationary points are: (-1, -1) and (-2, 14).