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victus00 [196]
3 years ago
8

Explain how to evaluate simplify and convert to radical form (x^3/8)^3/4

Mathematics
1 answer:
Vaselesa [24]3 years ago
3 0

Answer:

(x^\frac{3}{8})^\frac{3}{4}  = \sqrt[32]{x^9}

Step-by-step explanation:

Given

(x^\frac{3}{8})^\frac{3}{4}

Required

Convert to radical form

(x^\frac{3}{8})^\frac{3}{4}

Evaluate the exponents

(x^\frac{3}{8})^\frac{3}{4} = x^\frac{3*3}{8*4}

(x^\frac{3}{8})^\frac{3}{4} = x^\frac{9}{32}

Split the exponent

(x^\frac{3}{8})^\frac{3}{4} = (x^9)^\frac{1}{32}

Apply the following law of indices

(x^a)^\frac{1}{b} = \sqrt[b]{x^a}

So, we have:

(x^\frac{3}{8})^\frac{3}{4}  = \sqrt[32]{x^9}

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tresset_1 [31]

Answer:

See Below.

Step-by-step explanation:

We are given the graph of <em>y</em> = f'(x) and we want to determine the characteristics of f(x).

Part A)

<em>f</em> is increasing whenever <em>f'</em> is positive and decreasing whenever <em>f'</em> is negative.

Hence, <em>f</em> is increasing for the interval:

(-\infty, -2) \cup (-1, 1)\cup (3, \infty)

And <em>f</em> is decreasing for the interval:

\displaystyle (-2, -1) \cup (1, 3)

Part B)

<em>f</em> has a relative maximum at <em>x</em> = <em>c</em> if <em>f'</em> turns from positive to negative at <em>c</em> and a relative minimum if <em>f'</em> turns from negative to positive to negative at <em>c</em>.

<em>f'</em> turns from positive to negative at <em>x</em> = -2 and <em>x</em> = 1.

And <em>f'</em> turns from negative to positive at <em>x</em> = -1 and <em>x</em> = 3.

Hence, <em>f</em> has relative maximums at <em>x</em> = -2 and <em>x</em> = 1, and relative minimums at <em>x</em> = -1 and <em>x</em> = 3.

Part C)

<em>f</em> is concave up whenever <em>f''</em> is positive and concave down whenever <em>f''</em> is negative.

In other words, <em>f</em> is concave up whenever <em>f'</em> is increasing and concave down whenever <em>f'</em> is decreasing.

Hence, <em>f</em> is concave up for the interval (rounded to the nearest tenths):

\displaystyle (-1.5 , 0) \cup (2.2, \infty)

And concave down for the interval:

\displaystyle (-\infty, -1.5) \cup (0, 2.2)

Part D)

Points of inflections are where the concavity changes: that is, <em>f''</em> changes from either positive to negative or negative to positive.

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<em>f'</em> has extrema at (approximately) <em>x</em> = -1.5, 0, and 2.2.

Hence, <em>f</em> has inflection points at <em>x</em> = -1.5, 0, and 2.2.

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Answer:

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