Let f (x) = Using the centered 3-point formula for 1* derivatives: (x+1)2 f'(0) -2.04061 using h = 0.1 f'(0) N -2.01004 using h = 0.05 Applying Richardson's extrapolation on the above values, we get a better approximation of f'(0) which is 0 1.99985 -0.499998

Compute derivatives of the following functions. DO NOT SIMPLIFY YOUR ANSWERS. Explain which rules of differentiation you use for your computation.

1. (#11 Exam 1 Review) Find the equation of the line tangent to the graph of the function: f (x) = x-2x² at x = 1. You must use the Definition of the Derivative to obtain the derivative. 2. (#9 Exam 1 Review) Sketch the graph of the function g for which: g(0) = g(2) = g(4) = 0, g' (1) = g(3) = 0,g' (2) = -1, lim g(x) = ∞, and lim g(x) = -∞ x 5 x-1+ 3. State the three-part Definition of Continuity. 4. (#13 Exam 1 Review) The graph of f is given. State, with mathematically correct reasons, the values for which f is not differentiable. -2 VA 0 2 4 X (Description: Graph with three components. First component is horizontal line in the 2nd quadrant to x = -2. Sharp turn upwards in downward arc into the 1st quadrant and ends with open dot where x = 1. Second component is solid dot at x = 1 vertically below the open dot. Third component begins with open dot at x = 1 below the solid dot. The graph falls in downward arc shape into the 4th quadrant, takes a sharp turn at x = 3 and continues as…

Use backward difference approach of f(x) = sin x - x³ + 1 to estimate the first derivative of the function. Evaluate the derivative at x=1.4 or f'(1.4) using a step size of h=0.3 and h=0.15. Round-off answers to 6 decimal places. when h=0.30, f'(1.4) = -4.812947 when h=0.15, f'(1.4)= -5.251085 when h=0.30, f'(1.4)= -4.601757 when h=0.15, f'(1.4)= -5.139590 when h=0.30, f'(1.4)= -4.727838 when h=0.15, f'(1.4)= -5.206330 when h=0.30, f'(1.4)= -4.395859 when h-0.15, f'(1.4)= -5.029399 none of the choices