Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 − 32x2 + 3 (a) Find the interval 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 points. (x, y) = (smaller x-value) (x, y) = (larger x-value) 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.)

Answer:a) The increasing intervals would be from -4 to 0 and 4 to infinity. The decreasing interval would just be from negative infinity to -4 and 0 to 4.b) The local maximum comes at x = 0. The local minimums would be x = -4 and x = 4c) The inflection points are x= +/-√16/3Step-by-step explanation:To find the intervals of increasing and decreasing, we can start by finding the answers to part b, which is to find the local maximums and minimums. We do this by taking the derivatives of the equation. f(x) = x^4 - 32x^2 + 2f'(x) = 4x^3 - 64xNow we take the derivative and solve for zero to find the local max and mins. f'(x) = 4x^3 - 64x0 = 4x^3 - 64x0 = 4x(x + 4)(x - 4)x = -4 OR x = 4 OR 0Given the shape of a positive 4th power function function, we know that the first and last  would be a minimums and the second would be a maximum. As for the increasing, we know that a 4th power, positive function starts up and decreases to the local minimum. It also decreases after the local max. The rest of the time it would be increasing. In order to find the inflection point, we take a derivative of the derivative and then solve for zero. f'(x) = 4x^3 - 64xf''(x) = 12x^2 - 640 = 12x^2 - 6464 = 12x^216/3 = x^2+/- √16/3 = x-3/2 = x