Question #ae290

2 Answers
Dec 4, 2017

f(x)=tanx-cotx-x+1+pi/4f(x)=tanxcotxx+1+π4

Explanation:

f'(x)=sec^2x+cosec^2x-1

integrating

intf'(x)dx=intsec^2x+cosec^2x-1dx

f(x)=intsec^2xdx+intcosec^2xdx-int1dx

intsec^2xdx=tanx+c

intcosec^2xdx=-cotx+c

int1dx=x+c

f(x)=intsec^2xdx+intcosec^2xdx-int1dx=tanx-cotx-x+c

f(x)=tanx-cotx-x+c

finding the constant

when f(pi/4)=1

f(pi/4)=tan(pi/4)-cot(pi/4)-(pi/4)+c=1

f(pi/4)=1-1-(pi/4)+c=1

-(pi/4)+c=1

c=1+pi/4

now f(x) is

f(x)=tanx-cotx-x+1+pi/4

Dec 4, 2017

f(x)=tanx-cotx-x+pi/4+1

Explanation:

.

To find f(x) from f'(x), we have to take the integral of f'(x).

int(sec^2x+csc^2x-1)dx =intsec^2dx+intcsc^2xdx-intdx

f(x)=tanx-cotx-x+C

Now, we can plug in our initial conditions to solve for C:

f(pi/4)=tan(pi/4)-cot(pi/4)-pi/4+C=1

1-1-pi/4+C=1

cancelcolor(red)1-cancelcolor(red)1-pi/4+C=1

-pi/4+C=1

C=1+pi/4

f(x)=tanx-cotx-x+pi/4+1