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Use the second derivative test to classify the critical points.$$f(x)=3 x /\left(4 x^{2}+9\right)$$

$$\mathrm{m}(-3 / 2,-1 / 4), \mathrm{M}(3 / 2,1 / 4)$$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 3

Concavity and the Second Derivative

Derivatives

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04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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Use the second derivative …

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Classify the critical poin…

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question 44 would like you to use the second derivative test to classify the points of F of X equals three x, divided by four X squared plus nine. Using the quotient rule to find f prime of X that would be f prime of X is equal to four X squared, plus nine times three minus three x times eight x all over for X squared plus nine squared. Simplifying that it becomes negative. 12 X squared plus 27 over four x squared plus nine squared from there to find a critical point when you just set this equal to zero. So four X squared plus nine squared just gets multiplied by zero, which is zero so negative. 12 X squared plus 27 equals zero Solving for X that would just be plus or minus square root of 27/12, which is just negative. 3/2 and 3/2. Finding their critical points would just be plugging those numbers into F of X, so you have negative 3/2 negative 1/4 and 3/2 1/4 now to find out if they are concave up or down at those points you need F double prima bags, which would be using the quotient rule again on this function here, so that would be four X squared plus nine squared times Negative 24 x minus Negative 12 X squared plus 27 time 16 x times four X squared plus nine All divided by four x squared plus nine to the fourth. Now just simplifying. You can divide each of these sections by four X squared plus nine to the fourth, so F double prime of X is equal to negative. 24 x over four X squared plus nine Squared minus 16 X times Negative 12 X squared, plus 27 over four X squared plus nine to the third. Now, to get a common denominator, you need to multiply this side by four x squared plus 9/4 X squared plus nine. So after the prime of X, then becomes negative 24 x times four X squared plus nine minus 16 X times Negative 12 X squared plus 27 all over four X squared, plus nine to the third. If you multiply this out and then simplify it, you get F double prime of X is equal 2 96 x cubed minus 6 48 X Do I two by four x squared post nine to the third, which can be rewritten as 24 x times four X squared minus 27 over four X squared plus nine to the third. Now testing are critical points after a whole prime of 3/2 is negative 0.11 so concave down, it's a maximum and F double prime of negative 3/2 would be 0.11 which means this concave up or it's a minimum. So going back to a critical point, our maximum is at 3/2 1/4 and our minimum is at negative 3/2 negative 14

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