# How would you find the inflection point and the concavity of #g(x) = (5x - 2.6) / (5x - 6.76)^2#? I know I have to take the 2nd derivative but i'm not sure how because of the odd way this function is set up.?

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To find the inflection point and the concavity of ( g(x) = \frac{{5x - 2.6}}{{(5x - 6.76)^2}} ), you can follow these steps:

- Find the first derivative of ( g(x) ).
- Find the second derivative of ( g(x) ).
- Set the second derivative equal to zero and solve for ( x ) to find potential inflection points.
- Use the second derivative test to determine the concavity at these points.

Let's start by finding the first and second derivatives of ( g(x) ):

- ( g'(x) = \frac{{d}}{{dx}} \left( \frac{{5x - 2.6}}{{(5x - 6.76)^2}} \right) )
- ( g''(x) = \frac{{d^2}}{{dx^2}} \left( \frac{{5x - 2.6}}{{(5x - 6.76)^2}} \right) )

After finding the second derivative, you can proceed to find the potential inflection points and determine the concavity using the second derivative test.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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- How do you graph #f(x)=1/x-3x^3# using the information given by the first derivative?
- How do you use the first and second derivatives to sketch #f(x)= x^4 - 2x^2 +3#?
- What are the points of inflection, if any, of #f(x) =(x+4)/(x-2)^2#?

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