FIND THE X VALUES OF ALL POINTS, -2PI LESS THAN OR EQUAL TO X LESS THAN OR EQUAL TO 2PI, WHERE Y=X+B IS TANGENT TO Y=X+BSINX.
I KNOW THE ANSWER IS PI/2 AND 3PI/2, BUT WHY AREN'T THEY PLUS OR MINUS.
If they are tangent, they have the same coordinates, and the same slope.
y=x+b
y=x+bsinx
subtracting the first from the second..
0= b(sinx-1)
sinx=1 x=PI/2 or x=-3PI/2
slope:
both 1. The answers are pI/2, -3PI/2
To find the x-values of the points where the two equations y = x + b and y = x + bsin(x) are tangent to each other, you need to set the derivatives of the two equations equal to each other since the slope of a tangent line is equal to the derivative of the function at that point.
Let's start by finding the derivative of y = x + bsin(x):
The derivative of x is simply 1.
The derivative of bsin(x) with respect to x is bcos(x) (using the chain rule).
So the derivative of y = x + bsin(x) is 1 + bcos(x).
Now, let's find the derivative of y = x + b:
The derivative of x is 1.
The derivative of b with respect to x is 0 since b is just a constant.
So the derivative of y = x + b is simply 1.
Now, to find the x-values where these two derivatives are equal, set them equal to each other:
1 + bcos(x) = 1
Subtracting 1 from both sides, we get:
bcos(x) = 0
To solve for x, we look for values of x where the cosine of x equals zero. On the interval -2π ≤ x ≤ 2π, the cosine function reaches zero at x = π/2 and x = 3π/2.
Therefore, the x-values of the points where y = x + b is tangent to y = x + bsin(x) are π/2 and 3π/2.
As for why these values are not plus or minus, it's because when setting the two derivatives equal to each other and solving for x, we are finding the specific values of x where the slopes of the two functions are equal at that point. In this case, the slopes are equal at x = π/2 and x = 3π/2, not their negatives.