6. Determine a, b, c, and d so that the graph of y=ax^3+bx^2+cx+d has a point of inflection at the origin and a relative maximum at the point (2, 4). Sketch the graph.
y = ax^3 + bx^2 + cx + d
y' = 3ax^2 + 2bx + c
y'' = 6ax + 2b
y''(0) = 0
so b = 0
y'(2) = 0
so c = -12a
y(0) = 0
so d = 0
y = ax^3 - 12ax
y(2) = 4
so a = -1/4
y = -1/4 x^3 + 3x
y' = -3/4 x^2 + 3
y'' = -3/2 x
Well, determining the values of a, b, c, and d might be a bit challenging using humor alone. However, I can still assist you with the sketching part, as long as you promise not to laugh at my drawing skills!
So, let's start by discussing the point of inflection at the origin. A point of inflection occurs when the concavity of the curve changes. In other words, the graph is concave up on one side of the point and concave down on the other side.
Since we want the graph to have a point of inflection at the origin, we can say that the function's double derivative should change sign at x = 0. In other words, the second derivative of the function should be zero at x = 0.
Now, let's look at the relative maximum at (2, 4). At this point, the slope or derivative of the function should be zero, and the second derivative should be negative. This would indicate a maximum point.
Let's start by assuming a = 1 and b = 0 to simplify things. To make the graph go through (2, 4), we can plug in those values into the equation and find the corresponding value of c:
4 = (1)(2^3) + (0)(2^2) + c(2) + d
4 = 8 + 0 + 2c + d
d = -4
So far, we have a = 1, b = 0, c = -2, and d = -4. Let's plot these values on a rough sketch:
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Now, keep in mind that this is a rough sketch. With these values, the function should have a point of inflection at the origin and a relative maximum at (2, 4). You can see that the graph is concave up on the left side of the origin, then changes concavity and becomes concave down on the right side. The point (2, 4) should represent a relative maximum.
I hope this helps you visualize the graph a bit, even if my drawing skills are more clownish than accurate!
To determine the values of a, b, c, and d, we need to use the properties of the graph mentioned in the question.
1. Point of inflection at the origin (0, 0): This means that the second derivative of the function equals zero at x = 0.
2. Relative maximum at (2, 4): This means that the derivative of the function equals zero at x = 2.
Let's start by finding the second derivative of the function:
1. Find the first derivative of y = ax^3 + bx^2 + cx + d:
y' = 3ax^2 + 2bx + c
2. Find the second derivative:
y'' = 6ax + 2b
Since the point of inflection is at (0, 0), the second derivative must equal zero at x = 0:
6a(0) + 2b = 0
2b = 0
b = 0
Now we have the equation: y = ax^3 + cx + d.
To find the value of a, substitute x = 2 and y = 4 into the equation:
4 = a(2)^3 + c(2) + d
4 = 8a + 2c + d ----(1)
Since we need a relative maximum at (2, 4), the first derivative must equal zero at x = 2:
3a(2)^2 + 2c(2) + d = 0
12a + 4c + d = 0 ----(2)
From equations (1) and (2), we can set up a system of equations to solve for a, c, and d:
4 = 8a + 2c + d ----(1)
12a + 4c + d = 0 ----(2)
To simplify, let's multiply equation (2) by -1:
-12a - 4c - d = 0 ----(3)
Adding equations (1) and (3) together eliminates the d term:
4 = 8a + 2c + d
-12a - 4c - d = 0
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-8a - 2c = 4
Dividing both sides of this equation by -2:
4a + c = -2 ----(4)
Now we have two equations:
-8a - 2c = 4 ----(5) (from adding equations (1) and (3))
4a + c = -2 ----(4)
Solve equations (4) and (5) simultaneously to find the values of a and c.
Multiply equation (4) by 2:
8a + 2c = -4 ----(6)
Now add equations (5) and (6):
-8a - 2c = 4
8a + 2c = -4
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0 = 0
Since 0 = 0, this means that the values of a and c can be anything.
Lastly, find the value of d by substituting a = 0, b = 0, and c = -2 back into equation (1):
4 = 8(0) + 2(-2) + d
4 = -4 + d
d = 8
So the values of a, b, c, and d are:
a = any value
b = 0
c = -2
d = 8
Now we can sketch the graph. Please note that since we have the freedom to choose any value for "a," the shape of the graph can vary:
Graph of y = ax^3 + bx^2 + cx + d:
- The graph has a point of inflection at the origin (0, 0).
- The graph has a relative maximum at the point (2, 4).
- The graph passes through the point (0, 8) on the y-axis.
Please let me know if I can help you with anything else.
To determine the values of a, b, c, and d that satisfy the given conditions, we can use the information about the point of inflection and the relative maximum.
At the point of inflection (0, 0), the second derivative of the function will be zero. So, let's start by finding the second derivative of the function.
First, find the derivative of the function:
f(x) = ax^3 + bx^2 + cx + d
f'(x) = 3ax^2 + 2bx + c
Then, find the second derivative:
f''(x) = 6ax + 2b
Since the point of inflection is (0, 0), the second derivative at x = 0 should be zero:
f''(0) = 6a(0) + 2b = 0
2b = 0
b = 0
Now, we have:
f(x) = ax^3 + cx + d
f'(x) = 3ax^2 + c
Next, let's deal with the relative maximum at the point (2, 4). At that point, the derivative f'(x) should be zero.
f'(2) = 3a(2)^2 + c = 0
12a + c = 0
We obtained two equations from the given information:
1. 2b = 0
2. 12a + c = 0
Solving these equations simultaneously will give us the values of a, b, c, and d. Let's start with equation 1:
From 2b = 0, we know that b = 0.
Now, substitute b = 0 into equation 2:
12a + c = 0
12a = -c
Next, let's express c in terms of a:
c = -12a
Now, with b = 0 and c = -12a, we can rewrite the equation:
f(x) = ax^3 - 12ax + d
Finally, we need to find the value of d. To do this, we use the given information that the relative maximum occurs at the point (2, 4).
Substitute x = 2 and f(x) = 4 into the equation:
4 = a(2)^3 - 12a(2) + d
4 = 8a - 24a + d
4 = -16a + d
Now, express d in terms of a:
d = 4 + 16a
Therefore, the equation of the function is:
f(x) = ax^3 - 12ax + 4 + 16a
To sketch the graph, choose specific values for a and plot the corresponding graph. Remember that different values of a will result in different shapes of the curve.