calculus math help

Question

hi there im really struggling with these questions

Suppose that f is continuous on [0, 6] and that the only solutions of the equation  f (x) = 3 are x = 1 and x = 5. If  f (4) = 5, then which of the following statements must be true?

(i)f(2)>3

(ii)f(0)<3

(iii)f(6)<3

Let f (x) = sin x. Which of the following would you use to calculate f '(π/2) using the definition of derivative (i.e., first principles)?

(A) lim h→0 cos(−π/2 + h) h

(B) lim h→0 cos(−π/2 + h) − 1 h

(C) lim h→0 sin(π/2 + h) h

(D) lim h→0 sin(−π/2 + h) + 1 h

(E) lim h→0 sin(π/2 + h) − 1 h

(F) lim h→0 cos(π/2 + h) h

(G) lim h→0 cos(π/2 + h) + 1 h

(H) lim h→0 sin(−π/2 + h) h

evalute the flowing limit

lim h-> 0

sqrt(68-6(x+h) - sqrt(68-6x)/h

Answer 1

For the first question, if f(x) = 3 only when x = 1 and x = 5, f(4) = 5, and f is continuous on [0, 6], then f(x) > 3 on all of the interval (1, 5) because if f(x) < 3 then the continuous curve would have to cross over y = 3 again by the intermediate value theorem. Therefore f(2) > 3. The other statements are not necessarily true.

The limit definition of the derivative f'(π/2) is the limit as h → 0 of [sin(π/2 + h) - sin(π/2)]/h which is the limit as h → 0 of [sin(π/2 + h) - 1]/h.

My advice for the last limit is to multiply and divide by the conjugate sqrt(68 - 6(x+h)) + sqrt(68 - 6x). Multiply the two numerators and get a factor of h to cancel from the denominator.