Question: Consider the equation below. f(x) = 2x3 + 3x2 - 120x Exercise (a) Find the interval on which f is increasing. Find the interval on

 Consider the equation below. f(x) = 2x3 + 3x2 - 120xExercise (a) Find the interval on which f is increasing. Find theinterval on which f is decreasing. Part 1 of 6 For f(x)= 2x3 + 3x2 - 120x, we have f' ( x )= This factors into x + ( x - 4).Find the localminimum and maximum values of f. The function f(X) changes from increasingto decreasing at x = 5. Therefore, f(5) = is a maximumSubmit lyou cannot come back) Find the inflection point. Find the intervalon which fis concave up. Find the interval on which f isconcave down. Part 1 of 3 We have f'(x) = 6x2 +6x - 120, so f"(x) = which equals 0 when (x, f(x))
=Consider the equation below. f(x) = 2x3 + 3x2 - 12x (a)Find the interval on which f is increasing. (Enter your answer ininterval notation.) Find the interval on which f is decreasing. (Enter youranswer in interval notation.) (b) Find the local minimum and maximum valuesof f. local minimum local maximum (c) Find the inflection point. (x, y ) = Find the interval on which f is concaveup. (Enter your answer in interval notation.) Find the interval on whichf is concave down. (Enter your answer in interval notation.)Consider the equationbelow. f(x) = 4x3 + 9x2 - 54x +4 (a) Find theintervals on which f is increasing. (Enter your answer using interval notation.)Find the interval on which f is decreasing. (Enter your answer using

Consider the equation below. f(x) = 2x3 + 3x2 - 120x Exercise (a) Find the interval on which f is increasing. Find the interval on which f is decreasing. Part 1 of 6 For f(x) = 2x3 + 3x2 - 120x, we have f' ( x ) = This factors into x + ( x - 4).Find the local minimum and maximum values of f. The function f(X) changes from increasing to decreasing at x = 5. Therefore, f(5) = is a maximum Submit lyou cannot come back) Find the inflection point. Find the interval on which fis concave up. Find the interval on which f is concave down. Part 1 of 3 We have f'(x) = 6x2 + 6x - 120, so f"(x) = which equals 0 when (x, f(x)) =Consider the equation below. f(x) = 2x3 + 3x2 - 12x (a) Find the interval on which f is increasing. (Enter your answer in interval notation.) Find the interval on which f is decreasing. (Enter your answer in interval notation.) (b) Find the local minimum and maximum values of f. local minimum local maximum (c) Find the inflection point. ( x, y ) = Find the interval on which f is concave up. (Enter your answer in interval notation.) Find the interval on which f is concave down. (Enter your answer in interval notation.)Consider the equation below. f(x) = 4x3 + 9x2 - 54x +4 (a) Find the intervals on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (b) Find the local minimum and maximum values of f. local minimum value local maximum value (c) Find the inflection point. ( x, y ) = Find the interval on which f is concave up. (Enter your answer using interval notation.) Find the interval on which f is concave down. (Enter your answer using interval notation.)Consider the equation below. f(x) = x4 - 8x2 + 2 (a) Find the interval on which f is increasing. (Enter your answer in interval notation.) Find the interval on which f is decreasing. (Enter your answer in interval notation.) (b) Find the local minimum and maximum values of f. local minimum local maximum (c) Find the inflection points. ( x, y ) = (smaller x-value) ( x, y ) = (larger x-value) Find the interval on which f is concave up. (Enter your answer in interval notation.) Find the interval on which f is concave down. (Enter your answer in interval notation.)Use the graph to state the absolute and local maximum and minimum values of the function. (Assume each point lies on the gridlines. Enter your answers as a commaseparated list. If an answer does not exist, enter DNE.) absolute maximum value x absolute minimum value DNE y local maximum value(s) X local minimum value(s) 3' lJ lJJ Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x) = x3 + 3x2 144x Find the critical numbers of the function. y - 2 g(y) = - y2 - 2y + 4 Part 1 of 3 For g(y) = - y - 2 y2 - 2y + 4' , we have g'(y) =Tutorial Exercise Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 2x3 - 3x2 - 36x+ 6, [-3, 4] Part 1 of 4 The absolute maximum and minimum values of f occur either at a critical point inside the interval or at an endpoint of the interval. Recall that a critical point is a point where f '(x) = 0 or is undefined. We begin by finding the derivative of f. f '( x ) =Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 4x3 - 6x2 - 72x + 3, [-3, 4] absolute minimum absolute maximum

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