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

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.

Hopefully this helps!

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) :}#

(Graphed using desmos.com)

To sketch the graph satisfying the given conditions, follow these steps: 1. Mark the point (3, 5) on the graph, as it's given that f(3) = 5. 2. Sketch a curve that ascends to the left of x = 3, indicating f'(x) > 0 when x < 3. 3. Draw a curve that descends to the right of x = 3, indicating f'(x) < 0 when x > 3. 4. Ensure the graph is continuous at x = 3 to maintain the function's continuity. That's the basic outline for sketching the graph satisfying the conditions.