Find the open interval on which the function is increasing or decreasing, and locate all relative extrema?

Question

This is a calculus question dealing with extrema. Here's the function: f(x)=x/(x+1)

I don't know, I'm having a mental block, any help would be appreciated.

Answer 1

The function increases where its derivative is positive.

It decreases where its derivative is negative.

It has extrema where the derivative is zero.

What is the derivative of f(x) = x(x+1)^-1?

I can go further, but it sounds like you're looking for a gentle push instead of a hard shove.

Answer 2

First take the by-made of the function: f'(x) = 6x^2 + 6x -12 Now set f'(x) = 0 and likewise observe there are no factors the place the function is undefined... in basic terms an undemanding polynomial that we are in a position to remedy with the aid of factoring 6x^2 + 6x - 12 = 0 divide out with the aid of 6 to simplify x^2 + one million -2 =0 (x+2)(x-one million) = 0 x = -2, or x = one million are the severe numbers. Now, we are in a position to apply those factors because of the fact the ends of durations of increasing / reducing areas of the function. We in basic terms would desire to envision the sign of the by-product we got here across on the two edge of those factors **inf is infinity era (-inf, -2) (-2, one million) (one million, +inf) try: -3 0 2 f'(try) + - + ______/___________/ as a result, function is increasing (-inf, 2) and (one million, +inf) and reducing (-2, one million), (notice, there's a optimal at -2, and a minimum at one million) and reducing

Answer 3

f(x)=x/(x+1)
= [(x + 1) - 1]/(x+1)
= 1 - 1/(x + 1)

1. When x = 0 f(0) = 1 - 1/1 = 0
ie it passes through O (0, 0)

2. It is undefined when x = -1 as 1/0 is undefined.
So vertical asymptote at x = -1

3. As x increases and decreases without bound
1/(x + 1) approaches 0
So 1 - 1/(x + 1) approaches 1
ie lim x → ±∞ {x/(x + 1)} → 1

So horizontal asymptote at f(x) = 1

4. f'(x) = -1/(x + 1)² <0 for all x

ie f(x) is always decreasing

ie f(x) is decreasing for (-∞, -1) and for (-1, ∞)

and f(x) is undefined at x = -1.

Also f'(x) never equals 0 so no stationary points
f'(x) → 0 as x → ±∞

5. f''(x) = 2/(x + 1)³

If x > -1, f''(x) > 0 so curve is concave up.
ie concave up for (-1, ∞)
If x < 1, f''(x) < 0 so curve is concave down.
ie concave down for (-∞, -1)

Also f''(x) never equals 0 so no points of inflexion

Answer 4

No calculus can be fit in your brain. y=x/(x+1), y’=(x+1-x)/(x+1)^2=1/(x+1)^2;
Thus y’ is never =0, which means no local extrema!

Answer 5

dont know