(a) Use both the first and second derivative tests to show thatf(x) = 9sinx has a relative minimum at x = 0.O First derivative test:f" (x) = 18cos?x.f' (0) = 18; if x is near 0 thenf" < 0 forx < 0 and f' > Oforx > 0, sorelative minimum atx = 0.Second derivative test:f" (x) =18sin? x,f" (0) = 18 > 0, so relative minimum at x = 0.-O First derivative test:/'(x) = 18sin 2x.f"(0) = 0; if x is near O thenf' < 0 for x < 0and f' > 0 for x > 0, sorelative minimum at x = 0.%3DSecond derivative test:f (x) = 18cos 2x, f (0) = 18 > 0, so relative minimum atx = 0.O First derivative test:/"(x) = sin 2x, f' (0) = 0; if x is near 0 thenf' < Ofor x < 0 and f' > 0 for x > 0, so relativeminimum at x = 0.Second derivative test:f (x) = 2cos 2x.f (0) = 2 > 0, so relative minimum at x = 0.O First derivative test:/" (x) = 9sin 2x.f" (0) = 0; if x is near O thenf' < 0 for x < Oand f' > 0 for x > 0, so relativeminimum at x = 0.Second derivative test:f" (x) = 18cos 2x,f" (0) = 18 > 0, so relative minimum atx = 0.O First derivative test: /" (x) = 18cos 2x. ' (0) = 0; if x is near 0 thenf" < 0 for x < 0 andf" > Oforx > 0., sorelative minimum atx = 0.Second derivative test:f" (x) = 18sin 2x,f" (0) = 18 > 0, so relative minimum at x = 0.eTextbook and MediaHintAssistance Used(b) Use both the first and second derivative tests to show that g(x) = 2tanx has a relative minimum at x = 0.O First derivative test: g' (x) = 4tan x sec 2x, g' (0) = 0; if x is near 0 then g' < 0 for x < Oand g' > 0 for x > 0, sorelative minimum at x = 0.Second derivative test: 8" (x) = 4sec 2x(sec 2x + 2 tan 2x), g" (0) = 4 > 0, so relative minimum atx = 0.O First derivative test: g' (x) = 4tan x, g'(0) = 0; if x is near Otheng' < 0 for x < 0 and g' > 0 for x > 0, so relativeminimum at x = 0.Second derivative test: g" (x) = 4sec?x, g" (0) = 4 > 0, so relative minimum at.x = 0.O First derivative test: g' (x) = tan x sec? x, g' (0) = 0; if x is near 0 then g' < 0 for x < 0and g' > Ofor x > 0, sorelative minimum atx = 0.Second derivative test: g" (x) = sec²x(sec²x + tan²x), g" (0) = 1 > 0, so relative minimum atx = 0.O First derivative test: g' (x) = - 4tan x sec²x, g' (0) = 0; if x is near Othen g' > 0 for x < 0 and g' < 0 forx > 0, sorelative minimum atx = 0.Second derivative test: g" (x) = 4sec?x(sec²x – 2 tan²x), g" (0) = 4 > 0, so relative minimum at x = 0.O First derivative test: g'(x) = 4tan x sec²x, g' (0) = 0; ifx is near 0 then g' < O for x < Oand g' > 0for x > 0, sorelative minimum atx = 0.Second derivative test: g (x) = 4sec²x(sec²x+ 2 tan x), g (0) = 4 > 0, so relative minimum atx = 0. (c) Give an argument to explain without calculus why the functions in parts (a) and (b) have relative minima at.x = 0.O Both functions are squares of nonzero values when.x is close to 0 but.x + 0, and so are positive for values of x nearzero; both functions are zero at x = 0, so that must be a relative minimum.O Both functions are trigonometric functions, so they have a relative minimum at x = 0.O Both functions are zero at x = 0, so that must be a relative minimum.

As per our guidelines we are supposed to answer only one question i.e. the first one. For the…

(a) Use both the first and second derivative tests to show thatf(x) = 9sinx has a relative minimum at x = 0. O First derivative test:f" (x) = 18cos?x.f' (0) = 18; if x is near 0 thenf" < 0 forx < 0 and f' > Oforx > 0, so relative minimum atx = 0. Second derivative test:f" (x) = 18sin? x,f" (0) = 18 > 0, so relative minimum at x = 0. - O First derivative test:/"(x) = 18sin 2x.f' (0) = 0; if x is near Othenf" < O for x < 0 and f' > 0 for x > 0, so relative minimum at x = 0. %3D Second derivative test:f (x) = 18cos 2x, f (0) = 18 > 0, so relative minimum atx = 0. O First derivative test:/"(x) = sin 2x.f' (0) = 0; if x is near 0 thenf' < Ofor x < 0 and f' > 0 for x > 0, so relative minimum at x = 0. Second derivative test:f (x) = 2cos 2x.f (0) = 2 > 0, so relative minimum at x = 0. O First derivative test:/" (x) = 9sin 2x.f' (0) = 0; if x is near O thenf" < 0 for x < 0and f' > 0 for x > 0, so relative minimum at x = 0. Second derivative test:f (x) = 18cos 2x.f" (0) = 18 > 0, so relative minimum atx = 0. O First derivative test:f' (x) = 18cos 2x, f' (0) = 0; if x is near (0 thenf' < O for x < 0 and f' > Ofor x > 0, so %3D relative minimum atx = 0. Second derivative test:f (x) = 18sin 2x,f (0) = 18 > 0, so relative minimum at x = 0.

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...