2008 (2nd)Course2-4.1

We are to find a function f(x) such that its derivative is x^2+x-1 and the graph of y=f(x) is tangent to the straight line y=x+1.
First we find the coordinates of the point at which y=f(x) and y=x+1 are tangent.Since the slope of the tangent line is A,the x-coordinate of this point satisfies
x^2+x-B=0.
Therefore,the x-coordinate of this point is either CD or E,and hence the required point is either
(CD,FG) or (E,H).
Hence the required function f(x) is either
y=f(x)=1/Ix^3+1/Jx^2-x-K/L .......#1
or
y=f(x)=1/Ix^3+1/Jx^2-x+MN/O .......#2
Furthermore,since the graph of #1 can be obtained by translating the graph of #2 in the direction of the y-axis,the area of the region bounded by the graphs of #1 and #2 and the two straight lines x=CD and x=E is PQ/2.

To find the function f(x) that satisfies the given conditions, we need to break down the problem into several steps:

Step 1: Finding the point of tangency
We are given that the graph of y=f(x) is tangent to the line y=x+1. To find the coordinates of the point at which they are tangent, we need the slopes of both curves to be equal.

The slope of the tangent line y=x+1 is 1, so we want the derivative of f(x) to be equal to 1. We are given that the derivative of f(x) is x^2+x-1. Setting it equal to 1, we have:
x^2+x-1 = 1

Simplifying the equation, we get:
x^2+x-2 = 0

Now we can solve this quadratic equation to find the x-coordinate(s) of the point(s) of tangency.

Step 2: Determine the function f(x)
Since there are two possible x-coordinates (CD and E), there are two possibilities for the function f(x). Let's analyze both cases:

Case 1: Using x-coordinate CD
If the x-coordinate of the point of tangency is CD, we substitute this value into the equation for f(x):
y = f(x) = 1/Ix^3 + 1/Jx^2 - x - K/L (equation #1)

Case 2: Using x-coordinate E
If the x-coordinate of the point of tangency is E, we substitute this value into the equation for f(x):
y = f(x) = 1/Ix^3 + 1/Jx^2 - x + MN/O (equation #2)

Step 3: Finding the area bounded by the graphs of #1 and #2
The area of the region bounded by the graphs of #1 and #2, along with the lines x=CD and x=E, can be found by integrating the difference between #1 and #2 over the interval CD to E. The formula for the area of the region bounded by two curves is given by:
Area = ∫[CD, E] (|f(x) - g(x)|) dx

Here, f(x) is the function #1 and g(x) is the function #2. Integrating the absolute difference will give you the desired area.

Remember to determine the constants I, J, K, L, MN, and O using the given information or any additional constraints provided in the problem statement.

By following these steps, you should be able to find the function f(x) and calculate the area of the bounded region.

To find a function f(x) such that its derivative is x^2+x-1 and the graph of y=f(x) is tangent to the straight line y=x+1, we need to follow the steps below:

Step 1: Find the coordinates of the point where y=f(x) and y=x+1 are tangent.
The slope of the tangent line is equal to the derivative of f(x) at that point. Let's denote the slope as A. We can set up the equation:
x^2 + x - 1 = A

Step 2: Solve the equation from Step 1 to find the x-coordinate(s) of the tangent point.
By solving the equation x^2 + x - 1 = A, we can find the values of x.

Step 3: Use the x-coordinate(s) obtained in Step 2 to find the y-coordinate(s) of the tangent point(s).
Plug the x-coordinate(s) obtained in Step 2 into the equation y=f(x) to find the corresponding y-coordinate(s).

Step 4: Write the equation for the tangent line using the point(s) found in Step 3.
The equation for the tangent line passing through the point(s) (x,y) is y = x + 1.

Step 5: Set up an equation for the function f(x) using the slope and the point(s) from Step 3.
Using the slope of the tangent line A and the point(s) (x,y), we can set up the equation for f(x) as follows:
y = f(x) = 1/Ix^3 + 1/Jx^2 - x - K/L or
y = f(x) = 1/Ix^3 + 1/Jx^2 - x + MN/O

Step 6: Determine the relationship between the two possible equations for f(x) found in Step 5.
Since the graph of equation #1 can be obtained by translating the graph of equation #2 in the direction of the y-axis, we can say that the relationship between the two equations is a translation in the y-direction.

Step 7: Consider the area of the region bounded by the graphs of #1 and #2 and the two straight lines x=CD and x=E.
The area of the region bounded by the graphs of #1 and #2 and the two straight lines x=CD and x=E is given by PQ/2.

By following these steps, you will be able to find the function f(x) that satisfies the given conditions.