(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 =

The answer the given questions using the second derivative test.

(a) Use both the first and second derivative tests to show that f(x) = 5sin²xhas a relative minimum atx = 0. O First derivative test:f' (x) = 10sin 2x.f' (0) = 0; if x is near 0thenf' < 0 for x < 0andf' > O forx > 0, so relative minimum 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) = 10cos² x.f' (0) = 10; if x is near 0 thenf' < O forx < 0 and f' > O for x > 0, so relative minimum atx = 0. Second derivative test:f (x) = - 10sin°x.f" (0) = 10 > 0, so relative minimum atx = 0. O First derivative test:f'(x) = 5sin 2x,.f'(0) = 0; if x is near 0 thenf' < O for x < 0 and f' > O for x > 0, so relative %3D minimum at x = 0. Second derivative test:f (x) = 10cos 2x.f" (0) = 10 > 0, so relative minimum atx = 0. O First derivative test:f' (x) = 10cos 2x.f' (0) = 0; if x is near O thenf' < Oforx < 0 and f' > 0 forx > 0, so relative minimum atx = 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 thenf' < 0 for x < 0 andf' > Ofor x > 0, so relative %3D

(a) Use both the first and second derivative tests to show that f(x) = 5sin²xhas a relative minimum atx = 0. O First derivative test:f' (x) = 10sin 2x.f' (0) = 0; if x is near 0thenf' < 0 for x < 0andf' > O forx > 0, so relative minimum 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) = 10cos² x.f' (0) = 10; if x is near 0 thenf' < O forx < 0 and f' > O for x > 0, so relative minimum atx = 0. Second derivative test:f (x) = - 10sin°x.f" (0) = 10 > 0, so relative minimum atx = 0. O First derivative test:f'(x) = 5sin 2x,.f'(0) = 0; if x is near 0 thenf' < O for x < 0 and f' > O for x > 0, so relative %3D minimum at x = 0. Second derivative test:f (x) = 10cos 2x.f" (0) = 10 > 0, so relative minimum atx = 0. O First derivative test:f' (x) = 10cos 2x.f' (0) = 0; if x is near O thenf' < Oforx < 0 and f' > 0 forx > 0, so relative minimum atx = 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 thenf' < 0 for x < 0 andf' > Ofor x > 0, so relative %3D

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...

See similar textbooks

Related questions

Q: 1 x⁹ Determine the derivative of h. Let h(x) h'(x) = = = 1 9 X 10 Determine the slope of h at x = 2.…

A: h(x)=

Q: If the derivative of f is given by f'(x) =e* - 3x , at which of the following values of x does f…

A: Topic- application of derivatives

Q: 7) The graph below represents the derivative of the function f.That is, the graph isf'(x). Graph of…

A: As per our guidelines we are supposed to answer only first three parts.kindly repost others question…

Q: Use the definition of a derivative to find f (x) and f"(x). f(x) - 3x2 + 9x + 1 f(x) - F"(x) = Graph…

Q: 5. Suppose the derivative of a function f is f'(x) = (x - 2)²(x + 1)³(x − 8)¹. a) Find the regions…

Q: 1’) Let f be the function f(x) = x^2 +36/x^2 (a) Calculate f'(x) and find all stationary points for…

A: Solution - Given function - f(x)=x2+36x2 a.) f'(x)=2x+36-2x-2+1 f'(x)=2x-72x…

Q: Given the derivative f'(x), find the x-value for all relative extrema for the graph of the original…

A: Consider the given derivative. f'x=x2−5x+6 Put the first derivative equal to zero.…

Q: f'(x) = (2x – 1)ee".

Q: 12 Let. f(x) = ¹2. Please find f'(x), ƒ'(3), and the equation of the line tangent to f where x = 3.…

Q: Find the derivative of f(x) = xe-2x O (E) None of these O f'(x) = x²e-22 re (-2) O f'(x) = 2x²e-2r O…

Q: The first derivative of a continuous function y = f(x) is y' = x(x - 30)². Find y' and then use the…

Q: Find the derivative of the trigonometric function f(x)=x°t .5, °tanx. a. f'(x)=x°sec²x+ 5x*tanx b.…

A: Use the chain rule of derivative and simplify.

Q: Sketch the curve of f(x), a function whose first derivative has the following properties. You must…

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
%3D
minimum at x = 0.
Second derivative test:/" (x) = 10cos 2x.f" (0) = 10 > 0, so relative minimum at x
O First derivative test:f' (x) = 10cos²x,f'(0) = 10; if x is near () thenf' < O for x < 0 and f' > 0 for x > 0, so
0.
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 relative
minimum 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, so
relative 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 relative
minimum 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, so
relative 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, so
relative 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, so
relative 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
%3D
eTextbook 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.
%3D
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 near
0, 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 =

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

Step 2

Step 3

Step 4

Trending now

This is a popular solution!

Step by step

Solved in 4 steps

SEE SOLUTION Check out a sample Q&A here

Similar questions

To determine maximums and minimums by the Second Derivative Test, we differentiate y"=72 / (2-8)3 Substituting x = 14 into y'', _____ <,>, 0r = Substituting x = 2 into y'', ______ <,>, or =
6. Find f'(x). (1 point) Let f(x) = V 7x2 + 2 f' (æ) =
(7) Please show work as necessary.
2.(24) let g(v)