Consider the function f(x)=12x^5+30x^4–300x^3+2. f(x) has inflection points at (reading from left to right) x=D,E, and F. where D is___, E is____, and F is____.
The points of inflection are obtained by setting the second derivative of a function equal to zero
It is relatively easy to show that
f''(x) = 240x^3 + 360x^2 - 600
so
240x^3 + 360x^2 - 600 = 0
2x^3 + 3x^2 - 5 = 0
by a quick trial and error, I notice that x=1 satisfies the equation
so x-1 is a factor
dividing out I got
2x^3 + 3x^2 - 5 = 0
(x-1)(2x^2 + 5x + 5) = 0
so x=1 or x is equal to 2 complex roots.
Your function only has one point of inflection,
at x = 1
check your typing (or check my arithmetic)
Thanks for your help. It helped me figure out the correct answer. The second derivative is 240x^3+360x^2-1800x. From there I factored and got
120x(2x^2+3x-15)=0. then used the quadratic formula to get (-3+sqrt129)/4, (-3-sqrt129)/4, and 0.
To find the inflection points of the function f(x) = 12x^5 + 30x^4 - 300x^3 + 2, we need to find the values of x at which the concavity changes.
Step 1: Find the second derivative of f(x).
The inflection points can be found by analyzing the concavity of the function. To do this, we need to calculate the second derivative of f(x).
First, find the first derivative:
f'(x) = (d/dx) (12x^5 + 30x^4 - 300x^3 + 2)
= 60x^4 + 120x^3 - 900x^2
Next, find the second derivative:
f''(x) = (d^2/dx^2) (60x^4 + 120x^3 - 900x^2)
= 240x^3 + 360x^2 - 1800x
Step 2: Set f''(x) equal to zero and solve for x.
To find the points where the concavity changes, we need to solve the equation f''(x) = 0.
240x^3 + 360x^2 - 1800x = 0
Step 3: Factor out common terms.
Factor out x to simplify the equation:
x(240x^2 + 360x - 1800) = 0
Step 4: Solve for x.
Solve each factor separately:
Solving x = 0:
From x = 0, we obtain the inflection point at x = D = 0.
Solving 240x^2 + 360x - 1800 = 0:
Using the quadratic formula, we have:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
a = 240, b = 360, c = -1800
x = (-360 ± sqrt(360^2 - 4*240*(-1800))) / (2 * 240)
You can simplify and solve this equation to find the values of x at the inflection points E and F.
Therefore, the inflection points are:
D: x = 0
E: Solve for x using the quadratic equation
F: Solve for x using the quadratic equation