f(x) = 2 sin(x) + 2 cos(x), 0x 2 Exercise (a) Find the interval on which...

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f(x) = 2 sin(x) + 2 cos(x), 0≤x≤ 2π Exercise (a) Find the interval on which f is increasing. Find the interval on which f is decreasing. Step 1 For f(x) = 2 sin(x) + 2 cos(x), we have If this equals 0, then we have cos(x) = sin(x) tan(x) = 1 5π 4 X = f'(x) = 2 cos(x) - 2 sin(x) 2 cos(x) - 2 sin(x) Step 3 Step 2 If f'(x) is negative, then f(x) is decreasing increasing increasing If 0 < x < Step 4 If I <x< 4 5T 4 then f'(x) is positive 5л sin(x) ग Hence, in the interval 0 ≤ x ≤ 2π, f'(x) = 0 when x = 4 then f'(x) is negative which becomes I decreasing. If f'(x) is positive, then f(x) is positive and f(x) is increasing or negative, and f(x) is decreasing increasing decreasing. Step 5 If 5π < x < 2л, then f'(x) is positive 4 Step 6 Therefore, the interval on which f is increasing is (0,4).( 54,2x) 27 and the interval on which fis decreasing is π 5π 4' 4 Therefore, f(1) Exercise (b) Find the local minimum and maximum values of f. Step 2 = ---Select--- Step 1 π We know f(x) changes from increasing to decreasing at x = 4 is a maximum ✔ 2√2 (0,7). (7,2 2π 5п (77) 4' 4 local minimum value positive, and f(x) is increasing local maximum value (Enter your answer using interval notation.). 2√2 (Enter your answer using interval notation.) We know f(x) changes from decreasing to increasing at x = Therefore, f(5) 5л 4 So, the local minimum and maximum values of f are as follows. maximum increasing is a Step 2 5л We know f(x) changes from decreasing to increasing at x = 4 ---Select--- So, the local minimum and maximum values of f are as follows. Submit Skip (you cannot come back) local minimum value Step 1 We have f'(x) = 2 cos(x) - 2 sin(x), so f"(x) = X = local maximum value Exercise (c) Find the inflection points. Find the interval on which f is concave up. Find the interval on which f is concave down. which equals 0 when tan(x) = 3π 4 Submit Need Help? and x = Skip (you cannot come back) Read It Therefore, f(5) = Talk to a Tutor is a Hence, in the interval 0 ≤ x ≤ 2л, f"(x) = 0 when f(x) = 2 sin(x) + 2 cos(x), 0≤x≤ 2π Exercise (a) Find the interval on which f is increasing. Find the interval on which f is decreasing. Step 1 For f(x) = 2 sin(x) + 2 cos(x), we have If this equals 0, then we have cos(x) = sin(x) tan(x) = 1 5π 4 X = f'(x) = 2 cos(x) - 2 sin(x) 2 cos(x) - 2 sin(x) Step 3 Step 2 If f'(x) is negative, then f(x) is decreasing increasing increasing If 0 < x < Step 4 If I <x< 4 5T 4 then f'(x) is positive 5л sin(x) ग Hence, in the interval 0 ≤ x ≤ 2π, f'(x) = 0 when x = 4 then f'(x) is negative which becomes I decreasing. If f'(x) is positive, then f(x) is positive and f(x) is increasing or negative, and f(x) is decreasing increasing decreasing. Step 5 If 5π < x < 2л, then f'(x) is positive 4 Step 6 Therefore, the interval on which f is increasing is (0,4).( 54,2x) 27 and the interval on which fis decreasing is π 5π 4' 4 Therefore, f(1) Exercise (b) Find the local minimum and maximum values of f. Step 2 = ---Select--- Step 1 π We know f(x) changes from increasing to decreasing at x = 4 is a maximum ✔ 2√2 (0,7). (7,2 2π 5п (77) 4' 4 local minimum value positive, and f(x) is increasing local maximum value (Enter your answer using interval notation.). 2√2 (Enter your answer using interval notation.) We know f(x) changes from decreasing to increasing at x = Therefore, f(5) 5л 4 So, the local minimum and maximum values of f are as follows. maximum increasing is a Step 2 5л We know f(x) changes from decreasing to increasing at x = 4 ---Select--- So, the local minimum and maximum values of f are as follows. Submit Skip (you cannot come back) local minimum value Step 1 We have f'(x) = 2 cos(x) - 2 sin(x), so f"(x) = X = local maximum value Exercise (c) Find the inflection points. Find the interval on which f is concave up. Find the interval on which f is concave down. which equals 0 when tan(x) = 3π 4 Submit Need Help? and x = Skip (you cannot come back) Read It Therefore, f(5) = Talk to a Tutor is a Hence, in the interval 0 ≤ x ≤ 2л, f"(x) = 0 when