Question

How do you sketch the graph that satisfies f'(x)>0 when -1<x<3, f'(x)<0 when x<-1 and when x>3, and f(-1)=0, f(3)=4?

Answer 1

See sample drawing in explanation below. Example curve:
`f(x) = -x^3+3x^2+9x+2`

Explanation:

Sketching the graph is fairly straightforward once you interpret what each piece of the problem tells you about the shape of the graph.

`f'(x)>0` when `-1 < x < 3`

The first derivative tells you whether the graph is increasing ("going upwards") or decreasing ("going downwards"). If `f'(x)>0` (positive), then the graph is increasing. This statement tells us that from -1 to 3, the function will be increasing in y-values.

`f'(x)<0` when `x < -1` and `x > 3`

This statement says that `f'(x)<0` (negative) both to the left of -1 and to the right of 3. Visually, this means that the left edge of the number line to the left of -1 is going downward in y-values. On the right edge of the number line after 3, it again is falling in y-values.

`f(-1)=0` and `f(3)=4`

This gives us not just two points to draw that the graph goes through, but specifically the two points where the increasing/decreasing intervals change at.

On paper, on the left edge of the paper, draw the graph falling downwards from the upper-left corner of the paper until you hit the point `(-1,0)`, then curve it back upwards until you hit the point `(3,4)`, after which point make it fall downwards towards the lower-right corner of the paper.

Note: We aren't told anything about the second derivative, so we cannot be sure if the graph sections are concave up or down. We also don't know for sure what `f'(-1)` and `f'(3)` is (if even defined), so we don't know if the graph smoothly curves through those points or bounces sharply off.

Here's a possible example of what the graph might look like. For additional reading, look below the graph here for an example of how I derived this function for graphing.

graph{-x^3+3x^2+9x+2 [-6, 6, -16, 35]}

Example Function

Let's assume the function is a polynomial for convenience. Since we know that the left edge/tail of the graph points upward, and the right edge/tail of the graph points downward, the degree of the polynomial must be odd. ("End behavior of polynomials")

Further, there are two known critical points for the derivative (-1 and 3), which just so happen to be a minimum and maximum respectively. Thus, we know that `f'(-1)` and `f'(3)` are either 0 or undefined at those points. We can assume there are no further critical points for argument sake. We can also assume that the derivative is 0 at each for convenience sake.

This implies that -1 and 3 are zeroes of `f'(x)`, or put another way , that `(x+1)` and `(x-3)` are factors of `f'(x)`. Thus:

`f'(x) = A(x+1)(x-3)`, where `A` is some non-zero constant.

So:

`f'(x) = A(x^2-2x-3) = Ax^2-2Ax-3A`

However, we're told that `f'(x)<0` for `x<-1`. Let's examine `x = -2`:

`f'(-2) = A(-2)^2-2A(-2)-3A = 4A+4A-3A`

`= 5A < 0 => A < 0`

We know that `A < 0`. Continuing:

`f(x) = int f'(x) dx = int (Ax^2-2Ax-3A) dx`

`= A/3x^3 -Ax^2 -3Ax + C` where `C` is some constant.

Any choice of negative `A` and any `C` will define a graph that matches the criteria provided in this problem. I chose `A=-3` and `C = 2` to get `f(x) = -x^3+3x^2+9x+2` for my graph above.