Sketch a Graph of a Function that satisfies the following conditions

Question

f is continuous and odd

f'(x) = 2x if 0<x<2

f'(x) = -2/3 if 2<x<5

f'(x) = 0 if x>5

Answer 1

How about this:

Between 0 and 2 the function must be f(x) = x² + C if the derivative is 2x so we can just arbitrarily select C = 0.

Between 2 and 5 the function must be f(x) = -2/3x + C if the derivative is -2/3. To make it continuous with the first section we can raise it up by adding 16/3.

For x > 5 the function must be f(x) = C if the derivative is 0 and we can make C = 2 to make it continuous with the previous interval.

Since the function is odd, just flip it over to make the negative version.

Answer 2

Use the given information.

f(0) = 0 for all odd functions.

f(x) = x² if 0 < x < 2 (since the integration constant satisfies f(0) = 0).

f(x) = –2x/3 + 16/3 if 2 < x < 5 (choosing the integration constant to be continuous with x²)

f(x) = 2 if 5 < x (choosing the integration constant to be continuous with –2x/3 + 16/3).

Graph: https://www.wolframalpha.com/input/?i=Piecewise%5B%7B%7Bx²%2CAbs%5Bx%5D%3C2%7D%2C%7B-2Abs%5Bx%5D%2F3%2B16%2F3%2C2%3CAbs%5Bx%5D%3C5%7D%2C%7B2%2CAbs%5Bx%5D%3E5%7D%7D%5DSign%5Bx%5D