find the inflection points of x^4+5x^3-9x^2+6
An inflection point is where the second derivative of a function changes sign, also where curvature changes from concave to convex or vice versa.
Therefore to find inflection points, we only need to equate the second derivative of a function to zero, and solve for the zeroes.
f(x)=x^4+5x^3-9x^2+6
f"(x)=12x^2+30x-18
So solve for x where f"(x)=0.
Can you take it from here?
To find the inflection points of a function, you need to find the locations where the concavity of the graph changes. In other words, you are looking for points where the second derivative of the function changes sign.
Let's start by finding the second derivative of the given function.
Given function: f(x) = x^4 + 5x^3 - 9x^2 + 6
First derivative of f(x): f'(x) = 4x^3 + 15x^2 - 18x
Second derivative of f(x): f''(x) = 12x^2 + 30x - 18
Now, let's find the values of x where f''(x) changes sign. To do this, we need to solve the equation f''(x) = 0.
12x^2 + 30x - 18 = 0
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 12, b = 30, and c = -18.
Using the quadratic formula, we get:
x = (-30 ± √(30^2 - 4 * 12 * -18)) / (2 * 12)
x = (-30 ± √(900 + 864)) / 24
x = (-30 ± √1764) / 24
x = (-30 ± 42) / 24
Simplifying further, we have two potential solutions:
x1 = (-30 + 42) / 24 = 12/24 = 1/2
x2 = (-30 - 42) / 24 = -72/24 = -3
These are the values of x at which the concavity of the graph changes. So, the inflection points of the given function are (1/2, f(1/2)) and (-3, f(-3)).