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

Question

13139 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-02T16:38:47

We know the following.

•The function is increasing in the interval $(-oo, 3)$

•The function is decreasing in the interval $(3, oo)$

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

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

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

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

enter image source here

Hopefully this helps!

Answer 2 · 2017-01-02T17:02:58

Here are a couple more possibilities.

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

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

$f(x) = -abs(x-3)+5$ is shown below.

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

$f(x) = -(x-3)^(2/3) +5$ 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:

$f(x) = {(-(x-3)^2+1," if ",x < 3),(5," if ",x = 3),(-x+6," if ",3 < x) :}$

enter image source here

(Graphed using desmos.com)