As you can see in the image I made the example with n=6. In general, if the interval is divided into n equal parts we have that te right value of x in the kth rectangle (Rk) is
Rk = 2+ (1/2)k
So, f(Rk) = f(2+(1/2)k) = 2(2+(1/2)k)+1 = 4+k+1 = 5+k.
By "<span>f(x = 2x + 1" you likely meant "f(x) = 2x + 1."
The given interval is [2,5], and so the width of each interval is (5-2)/n, or (3/n). Since we are using right end points, the x value at the right endpoint of the kth rectangle is 2+k(3/n). Examples: if k=1, the value at the right endpoint is 2+1(3/n); if k=2, 2+2(3/n), and so on. If k=n, then the value at the right endpoint is 2+n(3/n), or 2+3=5, which agrees with the given interval [2,5].
Once again, the given function is f(x) = 2x + 1. At x = 2+k(3/n), the value of the function is 2[2+k(3/n)] + 1 (Answer). Check: If k=n, x = 2 + n(3/n) = 2+3 + 5, and f(5)=2(5) + 1 = 11.</span>
Sample Response: No, the graph is only increasing while the student rides his bike, rides the bus, and walks. It is stays the same while he waits for the bus and when the bus stops to let him off.
