What are the points of inflection, if any, of #f(x)=2x(2x-3)^3 #?

Question

Emily Ammerman · Accepted Answer

#(3/2,0)# is the only point of inflection

# f(x) = 2x(2x-3)^3 # graph{2x(2x-3)^3 [-2, 5, -11, 11]} Using the product rule and chain rule we have; # f'(x) = (2x)3(2x-3)^2(2) + (2)(2x-3)^3 # # :. f'(x) = 12x(2x-3)^2 + 2(2x-3)^3 # # :. f'(x) = (2x-3)^2(12x + 2(2x-3)) # # :. f'(x) = (2x-3)^2(12x + 4x - 6)) # # :. f'(x) = (2x-3)^2(16x - 6) # # :. f'(x) = 2(2x-3)^2(8x - 3) # .... [1] At critical points (min/max/poi) then # f'(x) = 0 # # f'(x) = 0 -> 2(2x-3)^2(8x - 3) = 0# # :. x=3/8,3/2 # To determine the nature of these points we need to examine the second derivative to see if f'(x) is increasing (min), decreasing (max) or 0 (inflection): Differentiating [1] wrt #x# we get; # f''(x) = (2(2x-3)^2)(8) + (2(2)(2x-3)(2))(8x - 3) # # :. f''(x) = 16(2x-3)^2 + 8(2x-3)(8x - 3) # When # x=3/8 => f''(3/8) = 16(6/8-3)^2 > 0 # (ie min) When # x=3/2 => f''(3/2) = 0 # (ie poi) When # x=3/2 => f(3/2) = 0 # Hence #(3/2,0)# is the only point of inflection

Adam Adkins · Answer

undefined To find the points of inflection of $ f(x) = 2x(2x-3)^3 $, you first need to find the second derivative of the function. Then, set the second derivative equal to zero and solve for $ x $. Finally, determine whether these solutions correspond to points of inflection by analyzing the concavity of the function.

First derivative: 
$$ f'(x) = 2(2x-3)^3 + 2x \cdot 3(2x-3)^2 $$

Second derivative:
$$ f''(x) = 6(2x-3)^2 + 2(2x-3)^2 + 12x(2x-3) $$

Setting the second derivative equal to zero:
$$ 6(2x-3)^2 + 2(2x-3)^2 + 12x(2x-3) = 0 $$

Solving this equation will give you the potential points of inflection. After finding these $ x $ values, you can determine whether they correspond to points of inflection by checking the concavity of the function around these points using the second derivative test.