Calculus Derivative / Limits help please?

Question

Hey, please help me do this, I have to submit for marks and I have no clue what to do.

A section road, represented by the line y = x + 2 with x =< 0 is to be connected to another section of road, represented by y = 2 - x with x >= 2, by means of a curved section of road, represented by a cubic curve y = ax^3 + bx^2 + cx + d, so that the resulting connected road has no sharp corners. Find the required values of a, b, c and d. In other words find a, b, c and d such that the function f(x) is differentiable, where:

f(x) = x + 2 where x =< 0
ax^3 + bx^2 + cx + d where 0 =< x =< 2
2 - x where x >= 2

Any help is greatly appreciated.
Thanks in advance.

Answer 1

y1(x) = x + 2 for x<=0
y2(x) = ax^3 + bx^2 + cx + d for 0<=x<=2
y3(x) = 2 - x for x>=2

The first thing one should realize is that y1(x) and y2(x) have the same y-intercept value; in other words, the functions share the point (0, 2). This fact can be used to find d:

y2(0) = 0 + 0 + 0 + d = 2 ----------> d = 2

It should also be apparent that y1(x) is tangent to y2(x) at the point (0, 2), and that y3(x) is tangent to y2(x) at point (2,0).
These facts are useful to know because of the following mathematical rule:

The slope of a tangent line is equal to the value of the other function's first derivative at the tangent point.

Slope of y1(x) = 1; Slope of y3(x) = -1

First derivative of y2(x) = y2'(x) = 3ax^2 + 2bx + c

Applying the above tangent rule to the 1st derivative, we get:

y2'(0) = 1 and y2'(2) = -1

y2'(0) = 0 + 0 + c = 1 -----------------> c = 1

y2'(2) = 12a + 4b + 1 = -1 ----------------> 12a + 4b = -2***

***Note: At this point, we have two variables left to solve for (a & b), so we only need one more relation involving these two values in order to get actual solutions.

We can get another equation with a & b by using one other observation: y2(x) and y3(x) both share the point (2,0)

y2(2) = a*(2)^3 + b*(2)^2 + (1)*(2) + 2 = 0

This simplifies to: 8a + 4b = -4

Multiply this 2nd relationship by -1, and add it to the prior one:

(12a + 4b = -2)
+ (-8a - 4b = 4)
----------------------
4a = 2 ------------------> a = 0.5

Plug this value back into either eqn to get b:

12*(0.5) + 4b = -2

4b = -8 -----------------------> b = -2


That's it!.....the equation for the curved section of road is:

y = 0.5x^3 - 2x^2 + x + 2


P.S. I graphed out all 3 eqns. on Excel, so if you'd like to see the results, let me know