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vesna_86 [32]
2 years ago
11

How do you do this problem?

Mathematics
2 answers:
levacccp [35]2 years ago
7 0

Answer:

x = -0.5

Step-by-step explanation:

I have marked on the graph the values of ƒ(x) that correspond to the different x-values.

At x = -2.5, ƒ(x) is decreasing, concave up, and > 0.  

At x = -1.5, ƒ(x) is decreasing, concave down, and < 0.  

At x = -0.5, ƒ(x) is increasing, concave up, and < 0.  

At x = 0.5, ƒ(x) is increasing, concave up, and > 0.  

At x = 2.5, ƒ(x) is decreasing, concave up, and > 0.  

The only point for which ƒ(x) is simultaneously increasing, concave up, and less than zero is x = -0.5.

exis [7]2 years ago
3 0
X=-.5, it’s less than 0, and concave up & increasing in this situation are the same.
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