The point of inflection for graph x^3/3-x^2-3x+7 I cannot seem to find it as stated.
For a cubic expression, there is one point of inflexion, namely where
f"(x)=0, or
2x-2=0
=> x=1
See following link for an image of the function:
http://imageshack.us/photo/my-images/819/1309284141.png/
but what would y be?
put that x into the function and determine y.
I got it thank you so much it makes much more sense when you help me.
To find the point(s) of inflection for a given graph, you need to follow these steps:
Step 1: Find the second derivative of the given function.
Step 2: Set the second derivative equal to zero and solve for x.
Step 3: Substitute the x-value(s) obtained from step 2 into the original function to find the corresponding y-value(s).
Let's go through each step in detail using the provided function:
Step 1: Find the second derivative of f(x) = x^3/3 - x^2 - 3x + 7
To find the second derivative, we differentiate the function twice. Starting with the first derivative:
f'(x) = d/dx (x^3/3 - x^2 - 3x + 7)
= (1/3)(d/dx)x^3 - d/dx(x^2) - d/dx(3x) + d/dx(7)
= (1/3)(3x^2) - 2x - 3
= x^2 - 2x - 3
Now, differentiate again to find the second derivative:
f''(x) = d/dx (x^2 - 2x - 3)
= 2x - 2
Step 2: Set the second derivative equal to zero and solve for x.
Set f''(x) = 0:
2x - 2 = 0
2x = 2
x = 1
Step 3: Substitute the x-value(s) obtained from step 2 into the original function to find the corresponding y-value(s).
To find the corresponding y-value(s), substitute x = 1 into the original function:
f(1) = (1)^3/3 - (1)^2 - 3(1) + 7
= 1/3 - 1 - 3 + 7
= 8/3
So, the point of inflection for the graph of f(x) = x^3/3 - x^2 - 3x + 7 is (1, 8/3).