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- X^2+2*x on the interval 0 to 3 with a step size of .5 using trapezoid ruleFind the maximum and minimum values of the function y=4x^(3)−4x+6 on the interval [0,2]. (Use decimal notation. Give your answer to three decimal places.)For function, find (a) the critical numbers; (b) the open intervals where the function is increasing; and (c) the open intervals where it is decreasing. y = x - 4 ln(3x - 9)
- For function, find (a) the critical numbers; (b) the open intervals where the function is increasing; and (c) the open intervals where it is decreasing. ƒ(x) = xe-3xQ4: Find and classify all critical points for the function: z2 = (x3/3)-(y3/3)+2xyConsider the function f(x)=2x^3−12x^2−30x+5 on the interval [−5,7]. Find the slope of the secant line (for this function) whose endpoints are (−5,f(−5)) and (7,f(7)). By the MVT, we know there exists a c in the open interval (−5,7) such that f′(c) is equal to slope of this secant line. Find the two values of c that work.
- Find the slope of the curve y=x^3-3 at the point P(-2,-11) by finding the limiting value of the slope of the secants through PThe function f (x, y) = 2x + 3y + 9 - x2 - xy - y2 has a maximum at some point (x, y). Find the values of x and y where this maximum occurs.At the end of the exam, all of a sudden your calculator breaks down. All you have left to do is to compute the value at (2.95,12.2) of the function f(x,y) = √xy, rounded to two decimals. You decide to put to work what you have learned in the math course: • You note that f (3, 12) = 6 and that (2.95, 12.2) is close to (3, 12).• You use the linear approximation of the function f at the point (3, 12) to compute an approxi- mate value for f (2.95, 12.2).• Is your rounded outcome on the exam correct?