# 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?

##### 1 Answer

See sample drawing in explanation below. Example curve:

#### Explanation:

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

The first derivative tells you whether the graph is increasing ("going upwards") or decreasing ("going downwards"). If

This statement says that

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

*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

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

This implies that -1 and 3 are zeroes of

So:

However, we're told that

We know that

Any choice of negative