(a) Use both the first and second derivative tests to show that f(x) = 5sinxhas a relative minimum at x = 0.First derivative test:f'(x) = 10sin 2x.f' (0) = 0; if x is near 0 then f' 0, so relative%3Dminimum at x = 0.Second derivative test:/" (x) = 10cos 2x.f" (0) = 10 0, so relative minimum at xO First derivative test:f' (x) = 10cos²x,f'(0) = 10; if x is near () thenf' 0, so0.relative minimum at.x = 0.- 10sin²x,f" (0) = 10 0, so relative minimum atx = 0.Second derivative test:f (x) =First derivative test: f'(x) = 5sin 2x, f' (0) = 0; if x is near () thenf' 0, so relativeminimum at x = 0.Second derivative test:f" (x) = 10cos 2x,f" (0) = 10 0, so relative minimum at x = 0.O First derivative test: f' (x) = 10cos 2x, f' (0) = 0; if x is near 0 then f' 0, sorelative minimum at.x = 0.Second derivative test:f" (x) = 10sin 2x,f" (0) = 10 0, so relative minimum at x = 0.O First derivative test: f' (x) = sin 2x. f' (0) = 0; if x is near 0 then f' 0, so relativeminimum at x = 0.Second derivative test:f" (x) = 2cos 2x, f" (0) = 2 0, so relative minimum atx = 0. (b) Use both the first and second derivative tests to show that g(x) = 5tanx has a relative minimum at x = 0.O First derivative test: g' (x) := 10tan x sec 2x, g' (0) = 0; if x is near 0 then g' 0, sorelative minimum atx = 0.Second derivative test: 8 (x) = 10sec 2x(sec 2x + 2 tan 2x), g (0) = 10 0, so relative minimum at x = 0.First derivative test: g' (x) = - 10tan x sec²x, g' (0) = 0; if x is near 0 then g' O for x 0, so relative minimum atx = 0.O First derivative test: g' (x) = tan x sec²x, g' (0) = 0; if x is near () then g' 0, sorelative minimum at.x = 0.Second derivative test: g (x) = sec?x(sec?x + tan?x),g" (0) = 1 0, so relative minimum at.x = 0.O First derivative test: g' (x) = 10tan x sec?x, g'(0) = 0; if x is near () then g' 0, sorelative minimum at.x = 0.Second derivative test: g" (x) = 10sec²x(sec²x + 2 tan?x), g"(0) = 10 0, so relative minimum at x = 0.O First derivative test: g' (x)minimum at x = 0.Second derivative test: g (x) = 10sec²x, g" (0) = 10 0, so relative minimum at x = 0.10tan x, g'(0) = 0; if x is near 0 then g' 0, so relative%3DeTextbook and Media(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 trigonometric functions, so they have a relative minimum atx = 0.%3DO Both functions are squares of nonzero values when x is close to 0 but x + 0, and so are positive for values of x near0, so that must be a relative minimum.0, so that must be a relative minimum.zero; both functions are zero atx =Both functions are zero at x =

Question

(a) Use both the first and second derivative tests to show that f(x) = 5sinxhas a relative minimum at x = 0.First derivative test:f'(x) = 10sin 2x.f' (0) = 0; if x is near 0 then f' < 0 for x < 0 andf' > Oforx > 0, so relative%3Dminimum at x = 0.Second derivative test:/" (x) = 10cos 2x.f" (0) = 10 > 0, so relative minimum at xO First derivative test:f' (x) = 10cos²x,f'(0) = 10; if x is near () thenf' < O for x < 0 and f' > 0 for x > 0, so0.relative minimum at.x = 0.- 10sin²x,f" (0) = 10 > 0, so relative minimum atx = 0.Second derivative test:f (x) =First derivative test: f'(x) = 5sin 2x, f' (0) = 0; if x is near () thenf' < O forx < 0 and f' > 0 for x > 0, so relativeminimum at x = 0.Second derivative test:f" (x) = 10cos 2x,f" (0) = 10 > 0, so relative minimum at x = 0.O First derivative test: f' (x) = 10cos 2x, f' (0) = 0; if x is near 0 then f' < 0 for x < 0 andf' > 0 for x > 0, sorelative minimum at.x = 0.Second derivative test:f" (x) = 10sin 2x,f" (0) = 10 > 0, so relative minimum at x = 0.O First derivative test: f' (x) = sin 2x. f' (0) = 0; if x is near 0 then f' < O for x < O and f' > Oforx > 0, so relativeminimum at x = 0.Second derivative test:f" (x) = 2cos 2x, f" (0) = 2 > 0, so relative minimum atx = 0. (b) Use both the first and second derivative tests to show that g(x) = 5tanx has a relative minimum at x = 0.O First derivative test: g' (x) := 10tan x sec 2x, g' (0) = 0; if x is near 0 then g' < 0 for x < 0 and g' > O forx > 0, sorelative minimum atx = 0.Second derivative test: 8 (x) = 10sec 2x(sec 2x + 2 tan 2x), g (0) = 10 > 0, so relative minimum at x = 0.First derivative test: g' (x) = - 10tan x sec²x, g' (0) = 0; if x is near 0 then g' > O for x < 0 and g' < Oforx > 0,so relative minimum at x = 0.Second derivative test: g (x) = 10sec²x(sec?x – 2 tan? x), g (0) = 10 > 0, so relative minimum atx = 0.O First derivative test: g' (x) = tan x sec²x, g' (0) = 0; if x is near () then g' < O for x < O and g' > O for x > 0, sorelative minimum at.x = 0.Second derivative test: g (x) = sec?x(sec?x + tan?x),g" (0) = 1 > 0, so relative minimum at.x = 0.O First derivative test: g' (x) = 10tan x sec?x, g'(0) = 0; if x is near () then g' < 0 for x < 0 and g' > 0forx > 0, sorelative minimum at.x = 0.Second derivative test: g" (x) = 10sec²x(sec²x + 2 tan?x), g"(0) = 10 > 0, so relative minimum at x = 0.O First derivative test: g' (x)minimum at x = 0.Second derivative test: g (x) = 10sec²x, g" (0) = 10 > 0, so relative minimum at x = 0.10tan x, g'(0) = 0; if x is near 0 then g' < 0 for x < Oand g' > 0 for x > 0, so relative%3DeTextbook and Media(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 trigonometric functions, so they have a relative minimum atx = 0.%3DO Both functions are squares of nonzero values when x is close to 0 but x + 0, and so are positive for values of x near0, so that must be a relative minimum.0, so that must be a relative minimum.zero; both functions are zero atx =Both functions are zero at x =

Accepted Answer

The answer the given questions using the second derivative test.