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Use the second derivative test to classify the critical points.$$f(x)=4 x^{3}-13 x^{2}+12 x+9$$

$$\mathrm{m}(3 / 2,45 / 4), \mathrm{M}(2 / 3,335 / 27)$$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 3

Concavity and the Second Derivative

Derivatives

Missouri State University

Oregon State University

Harvey Mudd College

University of Nottingham

Lectures

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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Find the critical points a…

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Find the critical point(s)…

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Classify the critical poin…

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question 41 would like you to use the second derivative test to classify the critical points of F of X equals four X cubed minus 13 X squared plus 12 x plus nine. Uh, first, to find those critical points, we need F prime of X. So a frame of X is equal to 12 x squared, minus 26 X plus 12 to find those critical points. That would be when F prime of X is equal to zero, and you can just use the quadratic formula for this so X equals 26 plus or minus square root 26 uh minus for times 12 times 12, all divided by two times 12. So X is equal to 26 plus or minus 10/24 which is just, um, 3/2 and 2/3. Plugging those points into ffx. You can get there y value, so it be 3/2 and 11.25 as well as to over three and 12.407 Now finding if those points are maximums or minimums. Take F double prime of X, which is equal 2 24 x minus 26 and plugging in these X values into F double prime of X at the old prime of 3/2 is 10, which means concave up, which is a minimum and F double prime of 2/3, is equal to negative 10, which is concave down, or a max so three over to 11 point 25 is a minimum, and to over 3 12 points. 407 is a max, and those are your answers to question 41.

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