Analyse the extreme curve sketching of the function f(x)=x3-3x.

To analyze the curve sketching of the function f(x) = x^2 - 3x, we will find its key features: the vertex (extreme point), intercepts, and concavity. Step 1: Find the vertex (extreme point). The x-coordinate of the vertex for a parabola f(x) = ax^2 + bx + c is given by x = -(b)/(2a). For f(x) = x^2 - 3x, we have a=1 and b=-3. x = -(-3)/(2(1)) = (3)/(2) Now, substitute x=(3)/(2) into the original function to find the y-coordinate: f((3)/(2)) = ((3)/(2))^2 - 3((3)/(2)) f((3)/(2)) = (9)/(4) - (9)/(2) f((3)/(2)) = (9)/(4) - (18)/(4) f((3)/(2)) = -(9)/(4) The vertex (extreme point) is ((3)/(2), -(9)/(4)). Since the coefficient of x^2 is positive (a=1), this vertex is a local minimum. Step 2: Find the intercepts. To find the y-intercept, set x=0: f(0) = (0)^2 - 3(0) = 0 The y-intercept is (0, 0). To find the x-intercepts, set f(x)=0: x^2 - 3x = 0 Factor out x: x(x - 3) = 0 This gives two x-intercepts: x = 0 or x - 3 = 0 x = 3 The x-intercepts are (0, 0) and (3, 0). Step 3: Determine intervals of increase and decrease. Find the first derivative of f(x): f'(x) = (d)/(dx)(x^2 - 3x) = 2x - 3 The function is increasing when f'(x) > 0: 2x - 3 > 0 2x > 3 x > (3)/(2) So, f(x) is increasing on the interval ((3)/(2), ). The function is decreasing when f'(x) < 0: 2x - 3 < 0 2x < 3 x < (3)/(2) So, f(x) is decreasing on the interval (-, (3)/(2)). Step 4: Determine concavity. Find the second derivative of f(x): f''(x) = (d)/(dx)(2x - 3) = 2 Since f''(x) = 2 > 0 for all x, the function is concave up everywhere. There are no inflection points. Summary of analysis for curve sketching: • The function is a parabola opening upwards. • The vertex (local minimum) is at ((3)/(2), -(9)/(4)). • The y-intercept is at (0, 0). • The x-intercepts are at (0, 0) and (3, 0). • The function is decreasing on (-, (3)/(2)) and increasing on ((3)/(2), ). • The function is concave up on (-, ). 3 done, 2 left today. You're making progress.