How do you determine the concavity for #f(x) = x^4 − 32x^2 + 6#?

#y'=4x^3-64x#

#y''=12x^2-64#,

#12x^2-64>0rArrx^2>64/12rArrx^2>16/3rArr#

#x<-4/sqrt3vvx>4/sqrt3#, or , better:

#x<-4/3sqrt3vvx>4/3sqrt3#. In this set the function has concavity up, in the complementary set it has concavity dawn, in #+-4/3sqrt3# there are two inflection points.

To determine the concavity of \( f(x) = x^4 - 32x^2 + 6 \), you need to find the second derivative of the function and then analyze its sign. First, find the first derivative of \( f(x) \) with respect to \( x \), denoted as \( f'(x) \). Then, find the second derivative of \( f(x) \), denoted as \( f''(x) \). After finding \( f''(x) \), examine where it is positive, negative, or zero. If \( f''(x) > 0 \) for a certain interval, the function is concave up on that interval. If \( f''(x) < 0 \) for a certain interval, the function is concave down on that interval. If \( f''(x) = 0 \) at a certain point, further analysis is required to determine the concavity at that point. Now, let's find the derivatives: \( f'(x) = 4x^3 - 64x \) \( f''(x) = 12x^2 - 64 \) To find where \( f''(x) \) is positive or negative, set it equal to zero and solve for \( x \): \( 12x^2 - 64 = 0 \) \( x^2 = \frac{64}{12} \) \( x^2 = \frac{16}{3} \) \( x = \pm \sqrt{\frac{16}{3}} = \pm \frac{4}{\sqrt{3}} \) As \( f''(x) \) is positive for \( x < -\frac{4}{\sqrt{3}} \) and \( x > \frac{4}{\sqrt{3}} \), the function \( f(x) \) is concave up on those intervals. And since \( f''(x) \) is negative for \( -\frac{4}{\sqrt{3}} < x < \frac{4}{\sqrt{3}} \), the function \( f(x) \) is concave down on that interval. Thus, the concavity of \( f(x) \) is as follows: Concave up: \( (-\infty, -\frac{4}{\sqrt{3}}) \) and \( (\frac{4}{\sqrt{3}}, +\infty) \) Concave down: \( (-\frac{4}{\sqrt{3}}, \frac{4}{\sqrt{3}}) \)