The function f has first derivative given by f'(x)=(x)^1/2/1+x+x^3. What is the x-coordinate of the inflection point of the grap

Question

Natalija [7]

3 years ago

15

The function f has first derivative given by f'(x)=(x)^1/2/1+x+x^3. What is the x-coordinate of the inflection point of the grap

h of f?
a. 1.008
b. 0.473
c. 0
d. -0.278
e. the graph has no inflection points

Mathematics

1 answer:

Anestetic [448] · Answer 1 · 2022-05-22T23:52:52+03:00

Inflection at f'(x) = 0

x^1/2 / (1 + x + x^3) = 0 

The x coordinate is x = 0 

or "C" from your choices there. 

The simplest way is to notice that this happens when x = 0. The reason is you got: 

x^1/2 / (some expression) 

at x = 0 the numerator is 0 so unless the denominator is also 0 the result is 0. 

The denominator for x = 0 is 1 so you get 

0/1 = 0.

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