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(b) Use both the first and second derivative tests to show that g(x) = 2tan²x 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' < 0for x < 0 and g' > 0 for x > 0, so relative minimum at x = 0. Second derivative test: g" (x) = 4sec 2x(sec 2x + 2 tan 2x). g" (0) = 4 > 0, so relative minimum at x = 0. O First derivative test: g' (x) = 4tan x, g' (0) = 0; if x is near 0 then g' < O for x < 0 and g' > 0 for x > 0, so relative minimum at x = 0. Second derivative test: g (x) = 4sec²x, g" (0) = 4 > 0, so relative minimum at x = 0. O First derivative test: g' (x) = tan x sec²x, g' (0) = 0; if x is near O then g' < Ofor x < 0 and g' > 0forx > 0, so relative minimum at x = 0. Second derivative test: g" (x) = sec²x(sec²x + tan²x), g¨ (0) = 1 > 0, so relative minimum at x = 0. O First derivative test: g' (x) = – 4tan x sec²x, g' (0) = 0; if x is near 0 then g' > O for x < 0 and g' < 0 for x > 0, so relative minimum at x = 0. Second derivative test: g (x) = 4sec²x(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; if x is near 0 then g' < O for x < 0 and g' > 0for x > 0, so relative minimum at x = 0. Second derivative test: g (x) 4sec²x(sec²x + 2 tan²x), g" (0) = 4 > 0, so relative minimum at x = 0.

(b) Use both the first and second derivative tests to show that g(x) = 2tan²x 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' < 0for x < 0 and g' > 0 for x > 0, so relative minimum at x = 0. Second derivative test: g" (x) = 4sec 2x(sec 2x + 2 tan 2x). g" (0) = 4 > 0, so relative minimum at x = 0. O First derivative test: g' (x) = 4tan x, g' (0) = 0; if x is near 0 then g' < O for x < 0 and g' > 0 for x > 0, so relative minimum at x = 0. Second derivative test: g (x) = 4sec²x, g" (0) = 4 > 0, so relative minimum at x = 0. O First derivative test: g' (x) = tan x sec²x, g' (0) = 0; if x is near O then g' < Ofor x < 0 and g' > 0forx > 0, so relative minimum at x = 0. Second derivative test: g" (x) = sec²x(sec²x + tan²x), g¨ (0) = 1 > 0, so relative minimum at x = 0. O First derivative test: g' (x) = – 4tan x sec²x, g' (0) = 0; if x is near 0 then g' > O for x < 0 and g' < 0 for x > 0, so relative minimum at x = 0. Second derivative test: g (x) = 4sec²x(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; if x is near 0 then g' < O for x < 0 and g' > 0for x > 0, so relative minimum at x = 0. Second derivative test: g (x) 4sec²x(sec²x + 2 tan²x), g" (0) = 4 > 0, so relative minimum at x = 0.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
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...
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(b) Use both the first and second derivative tests to show that g(x) = 2tan?x 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 < 0 and g' > 0 for x > 0, so relative minimum at x = 0. Second derivative test: 8" (x) = 4sec 2x(sec 2x + 2 tan 2x).g" (0) = 4 > 0, so relative minimum at x = 0. O First derivative test: g' (x) = 4tan x, g' (0) = 0; if x is near 0 then g' < O for x < 0 and g' > 0 for x > 0, so relative minimum at x = 0. Second derivative test: g (x) = 4sec²x, g" (0) = 4 > 0, so relative minimum atx = 0. O First derivative test: g' (x) = tan x sec?x, g'(0) = 0; if x is near 0 then g' < O forx < 0 and g' > 0 for x > 0, so relative minimum at x = 0. sec?x(sec?x + tan²x), g" (0) = 1 > 0, so relative minimum at x = 0. O First derivative test: g' (x) = – 4tan x sec² x, g' (0) = 0; if .x is near 0 then g' > 0 for x < 0 and g' < O for x > 0, so Second derivative test: g (x) = relative minimum at x = 0. Second derivative test: g" (x) = 4sec²x(sec?x – 2 tan?x), g" (0) = 4 > 0, so relative minimum at x = 0. O First derivative test: g' (x) = 4tan x sec²x, g' (0) = 0; if xis near 0 then g' < O for x < 0 and g' > 0 for x > 0, so relative minimum at x = 0. Second derivative test: g (x) 4sec?x(sec?x + 2 tan²x), g" (0) = 4 > 0, so relative minimum at x = 0. %3D
Transcribed Image Text:(b) Use both the first and second derivative tests to show that g(x) = 2tan?x 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 < 0 and g' > 0 for x > 0, so relative minimum at x = 0. Second derivative test: 8" (x) = 4sec 2x(sec 2x + 2 tan 2x).g" (0) = 4 > 0, so relative minimum at x = 0. O First derivative test: g' (x) = 4tan x, g' (0) = 0; if x is near 0 then g' < O for x < 0 and g' > 0 for x > 0, so relative minimum at x = 0. Second derivative test: g (x) = 4sec²x, g" (0) = 4 > 0, so relative minimum atx = 0. O First derivative test: g' (x) = tan x sec?x, g'(0) = 0; if x is near 0 then g' < O forx < 0 and g' > 0 for x > 0, so relative minimum at x = 0. sec?x(sec?x + tan²x), g" (0) = 1 > 0, so relative minimum at x = 0. O First derivative test: g' (x) = – 4tan x sec² x, g' (0) = 0; if .x is near 0 then g' > 0 for x < 0 and g' < O for x > 0, so Second derivative test: g (x) = relative minimum at x = 0. Second derivative test: g" (x) = 4sec²x(sec?x – 2 tan?x), g" (0) = 4 > 0, so relative minimum at x = 0. O First derivative test: g' (x) = 4tan x sec²x, g' (0) = 0; if xis near 0 then g' < O for x < 0 and g' > 0 for x > 0, so relative minimum at x = 0. Second derivative test: g (x) 4sec?x(sec?x + 2 tan²x), g" (0) = 4 > 0, so relative minimum at x = 0. %3D
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