x^3-6x^2+17x
Find the point of inflection of the graph of the function?
x,y ( , )
I tried to solve for it by getting the 1st derivative and setting it equal to 0 but i get sqrt,so am confused!
for the point of inflection you need the second derivative equal to zero
y = x^3 - 6x^2 + 17x
y' = 3x^2 - 12x + 17
y'' = 6x - 12
6x-12=0
x = 2
then y = 8 - 24 + 34 = 18
point of inflection is (2,18)
Thank you :)!
To find the point of inflection of a function, we need to determine the x-coordinate at which the concavity changes. In order to do this, we need to find the second derivative of the function and analyze its behavior.
Let's start by finding the first derivative of the function f(x):
f(x) = x^3 - 6x^2 + 17x
To find the first derivative, we can use the power rule:
f'(x) = 3x^2 - 12x + 17
Now, we need to find the second derivative by differentiating f'(x):
f''(x) = 6x - 12
Setting this second derivative equal to zero and solving for x will give us the x-coordinate of the inflection point:
6x - 12 = 0
Adding 12 to both sides:
6x = 12
Dividing both sides by 6:
x = 2
Now that we have found the x-coordinate of the inflection point, we need to find the corresponding y-coordinate by substituting x = 2 back into the original function:
f(2) = (2)^3 - 6(2)^2 + 17(2)
= 8 - 24 + 34
= 42
So, the point of inflection on the graph of the function is (2, 42).