Answer:
To find an equation that satisfies the given criteria, we can construct a piecewise defined function that consists of multiple polynomial segments. We will use calculus to ensure that the function meets the specified conditions.
Let's proceed step by step:
1. There are three inflection points:
To have three inflection points, we need at least three polynomial segments. Let's divide the function into four segments, each defined by a different polynomial.
2. There is at least one local maximum and one local minimum:
To ensure the existence of a local maximum and a local minimum, we can make the second polynomial segment have a local maximum and the third polynomial segment have a local minimum.
3. At least one critical point is not a given point:
To achieve this, we can make the first polynomial segment have a critical point that is not one of the given points.
4. The curve is continuous and differentiable throughout:
To ensure continuity and differentiability, we need to make sure that the polynomial segments join smoothly at the given points. This can be achieved by setting the derivatives of the adjacent polynomial segments equal at the given points.
Now, let's construct the piecewise defined function:
Segment 1: (1, 20) to (2, 4)
We can use a quadratic polynomial for this segment:
f1(x) = ax^2 + bx + c
Substituting the given points (1, 20) and (2, 4) into f1(x), we get two equations:
20 = a + b + c ---(1)
4 = 4a + 2b + c ---(2)
Solving equations (1) and (2), we find a = -3, b = 10, and c = 13.
So, f1(x) = -3x^2 + 10x + 13 for 1 ≤ x ≤ 2.
Segment 2: (2, 4) to (5, 3)
We can use a cubic polynomial for this segment:
f2(x) = dx^3 + ex^2 + fx + g
Substituting the given points (2, 4) and (5, 3) into f2(x), we get two equations:
4 = 8d + 4e + 2f + g ---(3)
3 = 125d + 25e + 5f + g ---(4)
Solving equations (3) and (4), we find d = -1/15, e = 1/3, f = -1/3, and g = 4.
So, f2(x) = (-1/15)x^3 + (1/3)x^2 - (1/3)x + 4 for 2 ≤ x ≤ 5.
Segment 3: (5, 3) to (6, 2)
We can use a quadratic polynomial for this segment:
f3(x) = hx^2 + ix + j
Substituting the given points (5, 3) and (6, 2) into f3(x), we get two equations:
3 = 25h + 5i + j ---(5)
2 = 36h + 6i + j ---(6)
Solving equations (5) and (6), we find h = -1/10, i = 1/10, and j = 3.
So, f3(x) = (-1/10)x^2 + (1/10)x + 3 for 5 ≤ x ≤ 6.
Segment 4: (6, 2) to (10, 1)
We can use a linear polynomial for this segment:
f4(x) = kx + l
Substituting the given points (6, 2) and (10, 1) into f4(x), we get two equations:
2 = 6k + l ---(7)
1 = 10k + l ---(8)
Solving equations (7) and (8), we find k = -1/4 and l = 5/2.
So, f4(x) = (-1/4)x + 5/2 for 6 ≤ x ≤ 10.
Putting it all together, the piecewise defined function that satisfies the given criteria is:
f(x) =
-3x^2 + 10x + 13 for 1 ≤ x ≤ 2
(-1/15)x^3 + (1/3)x^2 - (1/3)x + 4 for 2 ≤ x ≤ 5
(-1/10)x^2 + (1/10)x + 3 for 5 ≤ x ≤ 6
(-1/4)x + 5/2 for 6 ≤ x ≤ 10
This function has three inflection points, at least one local maximum and one local minimum, at least one critical point that is not a given point, and is continuous and differentiable throughout.
Explanation: