Studying functions
Increasing, decreasing, and extrema
Use the sign of the derivative to locate rises, falls, and possible turning points. Check candidates rather than treating every zero slope as an extremum.
1. Read derivative signs
1 / 4f=x^3-3x,\quad f\prime=3x^2-3 For x³−3x, f′=3(x²−1). It is positive outside [−1,1] and negative inside. The sign changes give a local maximum at −1 and local minimum at 1.
- f(a)
- -2
- f′(a)
- 0
The solid curve is f; the dashed line is its tangent at a. Read f(a) as height and f′(a) as slope. Questions use the input stated in their text.
2. A minimum at the vertex
2 / 4f=(x-1)^2,\quad f\prime=2(x-1) The nonnegative function (x−1)² reaches zero at x=1. Its derivative changes from negative to positive there, confirming a local and global minimum.
- f(a)
- 1
- f′(a)
- -2
The solid curve is f; the dashed line is its tangent at a. Read f(a) as height and f′(a) as slope. Questions use the input stated in their text.
3. Zero slope is not enough
3 / 4f=x^3,\quad f\prime=3x^2 For x³, f′(0)=0 but the derivative is positive on both sides. The graph keeps increasing through zero: there is no local maximum or minimum there.
- f(a)
- 1
- f′(a)
- 3
The solid curve is f; the dashed line is its tangent at a. Read f(a) as height and f′(a) as slope. Questions use the input stated in their text.
4. Include the endpoints
4 / 4f=x^3-3x,\quad -2\le x\le2 On [−2,2], evaluate x³−3x at −2, −1, 1, and 2. The values are −2, 2, −2, and 2. An absolute extreme can occur at an endpoint and need not be unique.
- a
- 1
- f(a)
- -2
Compare the heights at both endpoints and every interior critical point. The largest height is the absolute maximum; it can occur at more than one input.