Step 1 :Find the derivative of the function \(h(x) = 5x^3 - 15x\), which is \(h'(x) = 15x^2 - 15\).
Step 2 :Set the derivative equal to zero to find the critical points: \(15x^2 - 15 = 0\). Solving this equation gives the critical points \(x = -1\) and \(x = 1\).
Step 3 :Classify the critical points as relative maximum, minimum, or neither by checking the sign of the derivative on either side of these points. The point \((-1, -20)\) is a relative maximum and the point \((1, 20)\) is a relative minimum.
Step 4 :Find the second derivative of the function, which is \(h''(x) = 30x\).
Step 5 :Set the second derivative equal to zero to find potential points of inflection: \(30x = 0\). Solving this equation gives the potential point of inflection \(x = 0\).
Step 6 :Check the sign of the second derivative on either side of the potential point of inflection to confirm if it is a point of inflection. The point \((0, 0)\) is a point of inflection.
Step 7 :Determine the intervals where the function is increasing or decreasing. The function is increasing on the intervals \((-\infty, -1)\) and \((1, \infty)\), and decreasing on the interval \((-1, 1)\).
Step 8 :Determine the intervals where the function is concave up or concave down. The function is concave up on the interval \((0, \infty)\) and concave down on the interval \((-\infty, 0)\).
Step 9 :\(\boxed{\text{Final Answer: The relative maximum point is } (-1, -20) \text{ and the relative minimum point is } (1, 20). \text{ The function is increasing on the intervals } (-\infty, -1) \text{ and } (1, \infty), \text{ and decreasing on the interval } (-1, 1). \text{ The point of inflection is at } (0, 0). \text{ The function is concave up on the interval } (0, \infty) \text{ and concave down on the interval } (-\infty, 0).}\)