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Sketch the graph of the given function by demanding the appropriate information and points from...

Question:

Sketch the graph of the given function by demanding the appropriate information and points from the first and second derivatives:

{eq}y = 2x^3 - 24x - 7 {/eq}

a. What are the coordinates of the relative maxima?

b. What are the coordinates of the relative minima?

c. What are the coordinates of the points of inflection?

Relative Extremes:

The relative extremes of a function are critical points, we can find them by matching the first derivative to zero or when the first derivative does not exist. Inflection points are points where concavity changes.

Answer and Explanation:

We have the function:

{eq}f(x)= 2x^3 - 24x - 7 \\ {/eq}

Differentiating the function

{eq}f'(x)=6\,{x}^{2}-24 \\ {/eq}

{eq}f'(x)=0 {/eq} when {eq}x=2 \\ x=-2 \\ {/eq}

{eq}\begin{array}{r|D{.}{,}{5}} Interval & {-\infty<x<-2} & {-2<x<2} & {2<x<\infty } \\ \hline Test \space{} value & \ x=-5 & \ x=1 & \ x=3 \\ Sign \space{} of \ f'(x) & \ f'(-5)>0 & \ f'(1)<0 & \ f'(3)>0 \\ Conclusion & increasing & decreasing & increasing \\ \end{array} \\ {/eq}

Relative maximum at:

{eq}(-2, 25) \\ {/eq}

Relative minimum at: {eq}(2, -39) \\ {/eq}

Differentiating the function

{eq}f'(x)=6\,{x}^{2}-24 \\ f''(x)=12\,x \\ {/eq}

{eq}f''(x)=0 {/eq} when {eq}x=0 {/eq}

{eq}\begin{array}{r|D{.}{,}{5}} Interval & {-\infty<x<0} & {0<x<\infty} \\ \hline Test \space{} value & \ x=-1 & \ x=1 \\ Sign \space{} of \ f'' (x) & \ f'' (-1.5)<0 & \ f'' (1)>0 \\ Conclusion & concave \space down & concave \space up \\ \end{array} \\ {/eq}

Inflection point:

{eq}(0, -7) \\ {/eq}


Learn more about this topic:

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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
205K

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