Question

Find the open intervals on which the function is concave upward or concave​ downward, and find the location of any points of inflection.

Answer 1

To find a point of inflection, you have to have f''(x)=0(the double derivative)

f(x)=-4x^3+2x+2
f'(x)=-12x^2+2
f"(x)=-24x

Now we set f"(x) to 0

0=-24x

and get x=0 as a point of inflection, or more precisely (0,2) as a point of inflection(by plugging in 0 for f(x))

Now concaving upwards means that f"(x)>0 and concaving downwards means that f"(x)<0

Thus lets select a value to the right of the point of inflection and plug it into f"(x) to see the sign.

f"(1)=-24

Because f"(1)<0, x>0 is concaving downwards
since x>0 concaves downwards x<0 concaves upwards.

Thus we get (0,2) as a point of inflection, x>0 as concaving downwards, x<0 as concaving upwards.