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Flashcards: Use Riemann Sums

Let $f(x) = \frac{1}{7}x^{7} - \frac{6}{5} x^{5} + \frac{5}{3}x^{3} - 30x$ .

A relative maximum of the graph of can be located at:

$x=- \sqrt[4]{5}$

$x = \sqrt {6}$

$x= \sqrt[4]{5}$

$x =- \sqrt {6}$

The graph of has no relative maximum.

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Flashcards: Use Riemann Sums

Let $f(x) = \frac{1}{7}x^{7} - \frac{6}{5} x^{5} + \frac{5}{3}x^{3} - 30x$ .

A relative maximum of the graph of can be located at:

Your answer:

$x=- \sqrt[4]{5}$

Your answer:

$x = \sqrt {6}$

Your answer:

$x= \sqrt[4]{5}$

Your answer:

The graph of has no relative maximum.

Correct answer:

$x =- \sqrt {6}$

Explanation:

At a relative minimum (c,f(c)) of the graph f(x) , it will hold that f'(c) = 0 and f''(c) < 0 .

First, find f'(x) . Using the sum rule,

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{7}x^{7} - \frac{6}{5} x^{5} + \frac{5}{3}x^{3} - 30x \right )$

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{7}x^{7} \right )- \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{6}{5} x^{5} \right )+ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{5}{3}x^{3} \right )- \frac{\mathrm{d} }{\mathrm{d} x}\left ( 30x \right )$

Differentiate the individual terms using the Constant Multiple and Power Rules:

$f'(x) = \frac{1}{7} \cdot 7 x^{6} - \frac{6}{5} \cdot 5 x^{4} + \frac{5}{3} \cdot 3 x^{2} - 30x$

$f'(x) = x^{6} - 6 x^{4} +5 x^{2} - 30$

Set this equal to 0:

$x^{6} - 6 x^{4} +5 x^{2} - 30 = 0$

$(x^{6} - 6 x^{4} )+(5 x^{2} - 30) = 0$

$x^{4} (x^{2} - 6 )+ 5 (x^{2} - 6 ) = 0$

$(x^{4}+ 5 ) (x^{2} - 6 ) = 0$

Either:

$x^{4}+5 = 0$ , in which case, $x^{4} = -5$ ; this equation has no real solutions.

$x^{2}- 6=0$ has two real solutions, $-\sqrt {6}$ and $\sqrt {6}$ .

Now take the second derivative, again using the sum rule:

$f'(x) = x^{6} - 6 x^{4} +5 x^{2} - 30$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}(x^{6} - 6 x^{4} +5 x^{2} - 30)$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}(x^{6} )- \frac{\mathrm{d} }{\mathrm{d} x} ( 6 x^{4} )+\frac{\mathrm{d} }{\mathrm{d} x}( 5 x^{2} )- \frac{\mathrm{d} }{\mathrm{d} x}( 30)$

Differentiate the individual terms using the Constant Multiple and Power Rules:

$f''(x) =6 x^{5} - 6 ( 4 x^{3} )+ 5( 2x )- 0$

$f''(x) =6 x^{5} -2 4 x^{3} +10x$

Substitute $\sqrt {6}$ for :

$f''( \sqrt {6}) =6 ( \sqrt {6} )^{5} -2 4 ( \sqrt {6})^{3} +10( \sqrt {6})$

$f''(-\sqrt {6}) = 216\sqrt {6}-144 \sqrt {6} +10\sqrt {6}$

$f''(-\sqrt {6}) =82\sqrt {6} >0$

Therefore, has a relative minimum at $x = \sqrt {6}$ .

Now. substitute $-\sqrt {6}$ for :

$f''(-\sqrt {6}) =6 (-\sqrt {6} )^{5} -2 4 (-\sqrt {6})^{3} +10(-\sqrt {6})$

$f''(-\sqrt {6}) =- 216\sqrt {6}+144 \sqrt {6} -10\sqrt {6}$

$f''(-\sqrt {6}) =-82\sqrt {6} < 0$

Therefore, has a relative maximum at $x =- \sqrt {6}$ .

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Flashcards: Use Riemann Sums

Flashcards: Use Riemann Sums

Flashcards: Use Riemann Sums

Flashcards: Use Riemann Sums

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All Calculus AB Resources

Flashcards: Use Riemann Sums

Flashcards: Use Riemann Sums

Flashcards: Use Riemann Sums

Flashcards: Use Riemann Sums

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DMCA Complaint

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Complaint Details

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