# How do you sketch the graph that satisfies f'(x)>0 when x<3, f'(x)<0 when x>3#, and f(3)=5?

Jan 2, 2017

We know the following.

•The function is increasing in the interval $\left(- \infty , 3\right)$

•The function is decreasing in the interval $\left(3 , \infty\right)$

•The function passes through the point $\left(3 , 5\right)$

Considering the function is exclusively increasing in the interval $\left(- \infty , 3\right)$ and exclusively decreasing $\left(3 , \infty\right)$, then $\left(3 , 5\right)$ must be a maximum.

Any polynomial function of the form $- a {\left(x + 3\right)}^{n} + 5$, where $n$ is an even integer and , would satisfy the requirements given in the function.

The following is a graph of $- 2 {\left(x + 3\right)}^{6} + 5$, which satisfies perfectly the requirements of your question.

Hopefully this helps!

Jan 2, 2017

Here are a couple more possibilities.

#### Explanation:

HSBC244 has shown a nice graph that has derivative $f ' \left(3\right) = 0$.

Here are couple of graphs of functions that satisfy the requirements, but are not differentiable at $3$.

$f \left(x\right) = - \left\mid x - 3 \right\mid + 5$ is shown below.

graph{y = -abs(x-3)+5 [-14, 22.05, -6.16, 11.85]}

$f \left(x\right) = - {\left(x - 3\right)}^{\frac{2}{3}} + 5$ is shown on the next graph.

graph{-(x-3)^(2/3) +5 [-5.98, 14.025, -2.15, 7.844]}

For the unconventional here is a possible graph:

$f \left(x\right) = \left\{\begin{matrix}- {\left(x - 3\right)}^{2} + 1 & \text{ if " & x < 3 \\ 5 & " if " & x = 3 \\ -x+6 & " if } & 3 < x\end{matrix}\right.$

(Graphed using desmos.com)