For what values of x is #f(x)=(x-3)(x+2)(x-1)# concave or convex?

Given that: #f(x) =(x-3)(x+2)(x-1)# #:.# #f(x) =(x^2-x-6)(x-1)# #:.# #f(x) =(x^3-x^2-6x-x^2+x+6)# #:.# #f(x) =(x^3-2x^2-5x+6)#

For the function to be concave downward:#f''(x)<0# #f(x) =(x^3-2x^2-5x+6)# #f'(x) =3x^2-4x-5# #f''(x) =6x-4# For the function to be concave downward: #f''(x)<0# #:.##6x-4<0# #:.##3x-2<0# #:.## color(blue)(x<2/3) #

For the function to be concave upward:#f''(x)>0# #f(x) =(x^3-2x^2-5x+6)# #f'(x) =3x^2-4x-5# #f''(x) =6x-4# For the function to be concave upward: #f''(x)>0# #:.##6x-4>0# #:.##3x-2>0# #:.## color(blue)(x>2/3) #

To determine the concavity of the function \( f(x) = (x - 3)(x + 2)(x - 1) \), we need to find the second derivative of \( f(x) \) and examine its sign. The first derivative of \( f(x) \) is: \[ f'(x) = (x + 2)(x - 1) + (x - 3)(x - 1) + (x - 3)(x + 2) \] Expanding and simplifying, we get: \[ f'(x) = 3x^2 - 2x - 5 \] The second derivative of \( f(x) \) is: \[ f''(x) = \frac{d}{dx}(3x^2 - 2x - 5) = 6x - 2 \] To find where the function is concave or convex, we need to find the critical points by setting \( f''(x) \) equal to zero: \[ 6x - 2 = 0 \] \[ x = \frac{1}{3} \] Now, we can test the concavity: - For \( x < \frac{1}{3} \), \( f''(x) < 0 \), so \( f(x) \) is concave. - For \( x > \frac{1}{3} \), \( f''(x) > 0 \), so \( f(x) \) is convex. Thus, the function \( f(x) = (x - 3)(x + 2)(x - 1) \) is concave for \( x < \frac{1}{3} \) and convex for \( x > \frac{1}{3} \).