# 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?

Oct 13, 2017

See sample drawing in explanation below. Example curve:
$f \left(x\right) = - {x}^{3} + 3 {x}^{2} + 9 x + 2$

#### Explanation:

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

$f ' \left(x\right) > 0$ when $- 1 < x < 3$

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

$f ' \left(x\right) < 0$ when $x < - 1$ and $x > 3$

This statement says that $f ' \left(x\right) < 0$ (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.

$f \left(- 1\right) = 0$ and $f \left(3\right) = 4$

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 $\left(- 1 , 0\right)$, then curve it back upwards until you hit the point $\left(3 , 4\right)$, 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 $f ' \left(- 1\right)$ and $f ' \left(3\right)$ 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 $f ' \left(- 1\right)$ and $f ' \left(3\right)$ 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 $f ' \left(x\right)$, or put another way , that $\left(x + 1\right)$ and $\left(x - 3\right)$ are factors of $f ' \left(x\right)$. Thus:

$f ' \left(x\right) = A \left(x + 1\right) \left(x - 3\right)$, where $A$ is some non-zero constant.

So:

$f ' \left(x\right) = A \left({x}^{2} - 2 x - 3\right) = A {x}^{2} - 2 A x - 3 A$

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

$f ' \left(- 2\right) = A {\left(- 2\right)}^{2} - 2 A \left(- 2\right) - 3 A = 4 A + 4 A - 3 A$

$= 5 A < 0 \implies A < 0$

We know that $A < 0$. Continuing:

$f \left(x\right) = \int f ' \left(x\right) \mathrm{dx} = \int \left(A {x}^{2} - 2 A x - 3 A\right) \mathrm{dx}$

$= \frac{A}{3} {x}^{3} - A {x}^{2} - 3 A x + C$ 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 $f \left(x\right) = - {x}^{3} + 3 {x}^{2} + 9 x + 2$ for my graph above.