How do you find the inflection point, concave up and down for #f(x)=x^3-3x^2+3#?

Question

Ezra Adams · Accepted Answer

The first derivative is #f'(x)=3x^2-6x# and the second derivative is #f''(x)=6x-6=6(x-1)#. The second derivative is negative when #x<1#, positive when #x>1#, and zero when #x=1# (and of course changes sign as #x# increases "through" #x=1#). That means the graph of #f# is concave down when #x<1#, concave up when #x>1#, and has an inflection point at #x=1#. The coordinates of the inflection point are #(x,y)=(1,f(1))=(1,1)#.

Paisley Johnson · Answer

undefined To find the inflection point, concavity, and points of concavity change for $ f(x) = x^3 - 3x^2 + 3 $, follow these steps: 1. Find the second derivative of $ f(x) $: $ f''(x) = 6x - 6 $. 2. Set the second derivative equal to zero and solve for $ x $: $ 6x - 6 = 0 $ gives $ x = 1 $. 3. Use the second derivative test to determine the concavity at $ x = 1 $: - If $ f''(x) > 0 $ for $ x < 1 $, then the graph is concave up on that interval. - If $ f''(x) < 0 $ for $ x < 1 $, then the graph is concave down on that interval. - If $ f''(x) > 0 $ for $ x > 1 $, then the graph is concave up on that interval. - If $ f''(x) < 0 $ for $ x > 1 $, then the graph is concave down on that interval. 4. Substitute test values into $ f''(x) $ to determine the concavity: - $ f''(0) = -6 < 0 $, so the graph is concave down for $ x < 1 $. - $ f''(2) = 6 > 0 $, so the graph is concave up for $ x > 1 $. 5. Therefore, the inflection point is at $ x = 1 $, and the graph is concave down for $ x < 1 $ and concave up for $ x > 1 $.