Determine whether the statement is true or false. You do not...

Determine whether the statement is true or false. You do not have to justify your answer, just indicate true or false for each statement. a) If f'(x) = g'(x) for any x in the interval (2, 5), then f(x) = g(x) for any x in (2, 5). b) If f and g are increasing on an interval (a, b), then f – g is also increasing on (a, b). c) If f is continuous on [a, b], then g(x) = ?[a, x] f(t) dt is an antiderivative for f on [a, b]. d) If f'(a) = 0, then f has a local maximum or local minimum at a. e) If f and g are positive increasing functions on an interval (a, b), then fg is also increasing on (a, b). f) If f' is continuous on [1, 3] then ?[1, 3] f'(t) dt = f(3) – f(1). g) The antiderivative of 2x cos(x) is x² sin(x). h) If f is continuous on [3, 7], then ?[3, 7] xf(x) dx = x ?[3, 7] f(x) dx.

Transcript

00:01 For this problem, we want to determine whether the following statement is true or false.

00:05 Let's start with the first statement.

00:07 It says if f prime of x equals g prime of x in the interval 2 to 5, then f of x equals g of x for any x in the open interval 2 to 5.

00:17 Note that if f prime of x equals g prime of x, then if we take the antiderivative both sides, we have integral of f prime of x, dx equals the integral of g prime of x d x and from here we should get f of x plus c equals g of x plus d and because c and d are both arbitrary constants they may vary and may not be equal which implies that f of x is not necessarily equal to g of x so the first statement is false now for the second statement it says if f and g are increasing on an interval ab then f minus g is also increasing on ab.

01:08 So let's suppose f and g are increasing on the open interval ab.

01:14 If f and g are increasing on ab, then both f prime and g prime are greater than zero on this open interval ab.

01:22 Since the derivative with respect to x of f minus g is f prime minus g prime, then f prime minus g prime is greater than 0 if f prime is greater than g prime or f prime minus g prime less than 0 if f prime less than g prime.

01:50 So since there is a possibility for f minus g to be decreasing on the interval ab, then this statement is false.

01:59 And now for part c, if f is continual.

02:02 Us on the close interval ab, then g of x is equal to the integral from a to x, f of t, d, t, is an antiderivative for f on the closed interval ab.

02:14 If g of x is an antiderivative, then if you take the derivative both sides, g prime of x will be equal to the derivative with respect to x of the integral from a to x, f of t, dt, which is equal to f of x, by the fundamental theorem of calculus.

02:34 So this statement is true.

02:36 For part d, we have if f prime of a equals 0, f has local max or local min at a.

02:45 Now, if f prime of a equals 0, it doesn't not necessarily imply that f has a local max or min at x equals a, because sometimes our f is not changing its direction at x equals a...

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