For what values of x is #f(x)=(x-1)(x-3)(x+12)# concave or convex?

Question

Isabella Ackerman · Accepted Answer

Convex #x in ( - 12, 1 ) and ( 1, 3 )# and
concave #x notin [ - 12, 3 ]# Continuous and differentiable ( polynomial ) #f = x^3 ( 1 +O ( 1/x ) )#. As # x to +- oo, f to +- oo,# respectively. Zeros of f are #x= - 12, 1, 3. So, it is convex, for #x in ( - 12, 1 ) and( 1, 3 )#. See graph. graph{y-(x+12)(x-1)(x-2)=0[-20 20 -1000 1000]} Zoomed convex graph, for x in ( 1, 2 ): graph{y-(x+12)(x-1)(x-2)=0[1 2 -10 10]} In between successive leaving and returning to x-axis, the graph is convex. For that matter, #f = (x - x_1 ) ( x - x_2 )( x - x_3 )(x - x_n), x_i < x_(i+1) # is convex, #x in ( x_1, x_n)#, sans #x = x_i, i = 1, 2, 3, ..., n# and concave, for #x notin [ x_1, x_n ]#

Samantha Adkison · Answer

undefined To determine the concavity of the function $ f(x) = (x - 1)(x - 3)(x + 12) $, you need to examine the second derivative of the function. If the second derivative is positive, the function is concave up (convex), and if it is negative, the function is concave down. First, find the second derivative of $ f(x) $ by taking the derivative of the first derivative: $ f'(x) = 3x^2 - 22x - 35 $ $ f''(x) = 6x - 22 $ Now, set $ f''(x) $ equal to zero and solve for $ x $ to find the inflection points: $ 6x - 22 = 0 $ $ x = \frac{22}{6} = \frac{11}{3} $ Evaluate $ f''(x) $ at values less than and greater than $ x = \frac{11}{3} $ to determine the concavity: For $ x < \frac{11}{3} $: Choose $ x = 0 $ $ f''(0) = 6(0) - 22 = -22 < 0 $ For $ x > \frac{11}{3} $: Choose $ x = 4 $ $ f''(4) = 6(4) - 22 = 24 - 22 = 2 > 0 $ Therefore, the function $ f(x) = (x - 1)(x - 3)(x + 12) $ is concave down for $ x < \frac{11}{3} $ and concave up for $ x > \frac{11}{3} $.