can you please go over only the highlighted parts of the image. For example how did they get x= 20 and x= 180 for the first highlighted part? 3. Suppose that f(c) 2000 - .02³ +6²-2216r-1920. Find alllocal and global maxima and minima of this function using the firstand second derivative tests for maxima and minima. Plot the function.Is there a point of inflection? If so identify it and verify using the firstderivative test.1f'(x) =f"(x)= 2000.062² + 1224 -2216=-1122 +12Domain and range areboth R.So, look for optimia where t'=0-106x² +12x - 216~-0326² +62-108 = 0This quadratic bas 2 solution179180181x f'(x) f(x)192021-4.8304.774.77D4.81)- 4.8.-4.83f"20 when x2 = 100f" (99)f¹ (101) = 191197x=20x=1804= 191-97 => point ofin flection= 0First derivatie textf'No change insign of slope.1goes from at x = 20 =) minI f'3досfrom১০2 2150 =) такійdesivätheted.f" (20) >0 =) min.I f" (180) <0 =) maxim.f(x)2Second-216to co at20180

Given: The given function fx=2000x-0.02x3+6x2-2216x-1920. Our aim is to simplify the highlighted…

3. Suppose that f(x)=2000x-.02.r³+6²-2216r-1920. Find all local and global maxima and minima of this function using the first and second derivative tests for maxima and minima. Plot the function. Is there a point of inflection? If so identify it and verify using the first derivative test. f'(x) = = 2000.062² +1222-2216 f" (re) = =-1122 +12 1 Domain and range are both R. So, look for optimia where t"=0 -106x² +12x-216 x f'(x) f(x) 19 -4.83 20 0 21 4.77 179 180 ~-0326² +62-108 = 0 This quadratic bas 2 solutir 3 181 4.77 3 0 -4.83) f"2₂0 when x2 = 100 4.8 -4.8 x=20 x=180 f" (99) f¹ (101) = 191.97 = 191-97 => point of in flechon First derivatie test f' No change in sign of slope. 1 gues from at x = 20 = min. If' дос from so 22150 =) такій Second derivatreted. to so at I f" (20) >0 =) min. | f" (180) <0 =) maximu fix) 2 -216 20 180

3. Suppose that f(x)=2000x-.02.r³+6²-2216r-1920. Find all local and global maxima and minima of this function using the first and second derivative tests for maxima and minima. Plot the function. Is there a point of inflection? If so identify it and verify using the first derivative test. f'(x) = = 2000.062² +1222-2216 f" (re) = =-1122 +12 1 Domain and range are both R. So, look for optimia where t"=0 -106x² +12x-216 x f'(x) f(x) 19 -4.83 20 0 21 4.77 179 180 ~-0326² +62-108 = 0 This quadratic bas 2 solutir 3 181 4.77 3 0 -4.83) f"2₂0 when x2 = 100 4.8 -4.8 x=20 x=180 f" (99) f¹ (101) = 191.97 = 191-97 => point of in flechon First derivatie test f' No change in sign of slope. 1 gues from at x = 20 = min. If' дос from so 22150 =) такій Second derivatreted. to so at I f" (20) >0 =) min. | f" (180) <0 =) maximu fix) 2 -216 20 180

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter7: Analytic Trigonometry

Section7.6: The Inverse Trigonometric Functions

Problem 91E

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