This problem has multiple parts which are revealed sequentially after the preced

Question

This problem has multiple parts which are revealed sequentially after the preceding part has been answered correctly. Let f(x) = x + cos x on the interval [0, 2pi]. Fill in the blanks below. The curve y = f(x) has a y-intercept (is on the y-axis) at y = The curve y = f(x) has vertical asymptotes at x = The curve y = f(x) has horizontal asymptotes at y = f is increasing for x epsilon f is decreasing for x epsilon The local maxima of f occur at x = The local minima of f occur at x = The absolute maximum of f occurs at x = The absolute minimum of f occurs at x = f is concave upward for x epsilon f is concave downward for x epsilon The point(s) of inflection of f occur(s) at x =

Explanation / Answer

solution:

y intercept when x=0

y=0+cos0=1

hence y intercept is 1

2) vertical asymptotes occur where function is not defined

there is no point on domain where function is not defined hence no vertical asymptote

3) for finding horizontal asymptote we check

lim x-> infinity f(x) and it should be finite

as f(x) tends to infinite value when x tends infinity hence no horizontal asymptote

4) for increasing f'>0

1-sinx>0

sinx<1

hence it is always true except x=pi/2

hence interval

[0,pi/2) union (pi/2,2pi]

5) f'(x)<0

sinx>1

means no point in domain

6) now finding critical point

f'=0

1-sinx=0

sinx=1

x=pi/2

checking double derivate

f''=-cosx at pi/2

f''=0

now

checking endpoints

f(0)=1

f(2)=3

hence maxima 3 (6)

and minima 1 (7)

and point of inflection 0 (12)

and global maximaa and minima is same as local maxima and minima

f''=-cosx

for concave up

f''>0

-cosx>0

cosx<0

means pi/2 to 3pi/2

for concave down

f''<0

-cosx<0

cosx>0

means

0 to pi/2 and pi/2 to 3pi/2

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