What are the points of inflection, if any, of #f(x) = -x^6 -4x^5 +5x^4 #?

Question

What are the points of inflection, if any, of #f(x) = -x^6 -4x^5 +5x^4   #?

Evelyn Agard · Accepted Answer

The points of inflection are:
#(-3.2770, 849.8146) and (0.6103, 0.3033)# Points of inflection are points on the function where the second derivative changes sign, that is the graph goes from concave upward to concave downwards or vice-versa. Let #y = f(x)# We must find the second derivative of the function: #(d^2y)/(dx^2) = y''(x)# #y'(x) = -6x^5 - 20x^4 + 20x^3#, using the power rule Then, #y''(x) = -30x^4 - 80x^3 + 60x^2# Let us find where the second derivative is zero, to find the critical x-values. #-30x^4 - 80x^3 + 60x^2 = 0# #-10x^2(3x^2 + 8x - 6) = 0# #x=0# is a solution here, and to find the other two solutions, we solve the following equation: #3x^2 + 8x - 6 = 0# Using the quadratic formula, #x = (-8 ± sqrt(64 - (-72)))/(6) = (-8 ± sqrt(136))/(6)# #x approx -3.27698396495, or x approx 0.610317298282# Let us find the concavity of #f(x)# on these intervals: #on (-infty, -3.27698396495), f''(x)<0# #on (-3.27698396495, 0), f''(x)>0# #on (0, 0.610317298282), f''(x)>0# #on (0.610317298282, infty), f''(x)<0# So in succession on #(-infty, infty)#, the sign line looks like this: #-, +, +, -# Therefore, #f''(x)# only changes signs at the x-values of #-3.27698396495# and #0.610317298282# Since #f''(x)# changes signs at these values, #f(x)# has inflection points here as well. The coordinates of the inflection points are: #(-3.27698396495, f(-3.27698396495))# and #(0.610317298282, f(0.610317298282))# Therefore, the coordinates are, rounding to 4 decimal places: #(-3.2770, 849.8146) and (0.6103, 0.3033)#

Emily Ammerman · Answer

undefined To find the points of inflection of $ f(x) = -x^6 - 4x^5 + 5x^4 $, we first need to find the second derivative of the function and then solve for where it equals zero.

First derivative of $ f(x) $:
$$ f'(x) = -6x^5 - 20x^4 + 20x^3 $$

Second derivative of $ f(x) $:
$$ f''(x) = -30x^4 - 80x^3 + 60x^2 $$

To find the points of inflection, we need to solve for $ x $ where $ f''(x) = 0 $:

$$ -30x^4 - 80x^3 + 60x^2 = 0 $$

Factoring out $ -10x^2 $ from each term:

$$ -10x^2(3x^2 + 8x - 6) = 0 $$

Using the quadratic formula to solve $ 3x^2 + 8x - 6 = 0 $:

$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

Where $ a = 3 $, $ b = 8 $, and $ c = -6 $:

$$ x = \frac{{-8 \pm \sqrt{{8^2 - 4 \cdot 3 \cdot (-6)}}}}{{2 \cdot 3}} $$
$$ x = \frac{{-8 \pm \sqrt{{64 + 72}}}}{{6}} $$
$$ x = \frac{{-8 \pm \sqrt{{136}}}}{{6}} $$
$$ x = \frac{{-8 \pm 2\sqrt{{34}}}}{{6}} $$
$$ x = \frac{{-4 \pm \sqrt{{34}}}}{{3}} $$

So, the points of inflection are $ x = \frac{{-4 + \sqrt{{34}}}}{{3}} $ and $ x = \frac{{-4 - \sqrt{{34}}}}{{3}} $.