For f''(x) = (x-6)^5 (x-2)^3 (x+5)^7 a) f(x) has an inflection point at x = -5 and has an...

Question:

For {eq}f''(x) = (x-6)^5 (x-2)^3 (x+5)^7 {/eq}

a) {eq}f(x) {/eq} has an inflection point at {eq}x = -5 {/eq} and has an inflection point at {eq}x = 2 {/eq}. Explain why that statement is true.

b) Also, what is the rule that will determine if {eq}f'(x) {/eq} has a maximum/minimum ?

Inflection Points and Maximum/Minimum of a Function


Given the second derivative of a function f(x) we comment on its inflection points. Then we state the criteria necessary for the function to have a relative maximum and relative minimum at points(s) in its domain. The concepts are those from Calculus and involve the first and second derivatives as well as zeros of functions.


Answer and Explanation:


a) Since {eq}f''(-5)=0 {/eq} and {eq}f''(2)=0, {/eq} therefore {eq}x=-5 \; {\rm and} \; x=2 {/eq} are both inflection points for {eq}f(x). {/eq}


b) From Calculus, {eq}f'(x) {/eq} will have a relative maximum at point {eq}x=a {/eq} in its domain if {eq}f''(a)=0 \; {\rm and} \; f'''(a)<0. {/eq}

Similarly, {eq}f'(x) {/eq} will have a relative minimum at point {eq}x=a {/eq} in its domain if {eq}f''(a)=0 \; {\rm and} \; f'''(a)>0. {/eq}

There is another way a function can have a relative maximum or minimum at a point as follows:

{eq}f'(x) {/eq} may have a relative maximum or relative minimum at point {eq}x = a {/eq} in its domain if {eq}f''(a) {/eq} does not exist or is UNDEFINED. In that case one would need to use graphical techniques to determine whether a relative maximum or relative minimum exists at this point. This, since the second derivative test will no longer work owing to higher order derivatives NOT defined either.


Learn more about this topic:

Loading...
Using Differentiation to Find Maximum and Minimum Values

from Math 104: Calculus

Chapter 9 / Lesson 4
48K

Related to this Question

Explore our homework questions and answers library