Help with Calculus problem

Question

a) Find f.

f '(t) = sec(t)(sec(t) + tan(t)), − 𝜋/2 < t < 𝜋/, f (𝜋/4) = −7

f(t) = ???

b) Find f.

f ''(x) = x^-2, x > 0, f(1) = 0, f(6) = 0

f(x) = ???

 

Answer 1

a) f'(t)= sec(t)(sec(t) + tan(t))

= sec^2(t) + sec(t)tan(t)

"-pi/2<t< pi/" ? (maybe <pi/2?)

f(pi/4)=-7

f(t)= tan(t} +sec(t) + c

f(pi/4) = -7 = tan(pi/4) + sec(pi/4) + c = 1 + sqr(2) + c

c = -7 -1 -sqr2

f(t) = tan(t) + sec(t) -8- sqr(2)

b) f"(x)=x^-2 = 1/x^2

f'(x)= -x^-1 + c where c is an unknown constant

f'(1)=0= -(1)^-1 +c (that "f(1)" must have really been "f'(1)"?

if so, then solve for c as follows:)

0 =-1 +c

c=1

f'(x)= -1/x +1

f(x)= -lnx+x+ k where k is another unknown constant

f(6)=0= -ln6 +6+k

k= ln6 - 6

f(x) = -lnx +x +ln6 -6= ln(6/x) +x -6

f(x)= ln(6/x) +x-6