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3 years ago

Which graph represents an odd function?

Mathematics

2 answers:

Nookie1986 [14]3 years ago

8 0

Answer:

<h2>The first graph in the second image is an odd function.</h2>

Step-by-step explanation:

An odd function has a graph that it's symmetric about the origin, that is, the origin is like a mirror. In other words, the graph of an odd function has a specific symmetry about the origin.

So, we have to look for those graph that has symmetrical points in opposite quadrants, I and III or II and IV.

You can observe that the first graph of the second image has this behaviour. You can see that the points are symmetrical across the origin. If you graph a line defined as y=-x, you will observe that such line acts like a mirror.

Therefore, the odd function is the first graph in the second image.

dem82 [27]3 years ago

7 0

Answer:

The first graph in the second image is an odd function.

Step-by-step explanation:

An odd function has a graph that it's symmetric about the origin, that is, the origin is like a mirror. In other words, the graph of an odd function has a specific symmetry about the origin.

So, we have to look for those graph that has symmetrical points in opposite quadrants, I and III or II and IV.

Therefore, the odd function is the first graph in the second image.

You might be interested in

Find the derivative of following function.

Aleks04 [339]

Answer:

$\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}$

General Formulas and Concepts:
Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]:
$\displaystyle (cu)' = cu'$

Derivative Property [Addition/Subtraction]:
$\displaystyle (u + v)' = u' + v'$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
$\displaystyle (uv)' = u'v + uv'$

Derivative Rule [Quotient Rule]:
$\displaystyle \bigg( \frac{u}{v} \bigg)' = \frac{vu' - uv'}{v^2}$

Derivative Rule [Chain Rule]:
$\displaystyle [u(v)]' = u'(v)v'$

Step-by-step explanation:

*Note:

Since the answering box is way too small for this problem, there will be limited explanation.

Step 1: Define

Identify.

$\displaystyle y = \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \times \frac{\tan^2 x + 5x}{\csc^2 x + 3}$

Step 2: Differentiate

We can differentiate this function with the use of the given derivative rules and properties.

Applying Product Rule:

$\displaystyle y' = \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) '$

Differentiating the first portion using Quotient Rule:

$\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( \cos^2 x - 3\sqrt{x} + 6 \big)' \big( \sin^2 x + 6 \big) - \big( \sin^2 x + 6 \big)' \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}$

Apply Derivative Rules and Properties, namely the Chain Rule:

$\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}$

Differentiating the second portion using Quotient Rule again:

$\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( \tan^2 x + 5x \big)' \big( \csc^2 x + 3 \big) - \big( \csc^2 x + 3 \big)' \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Apply Derivative Rules and Properties, namely the Chain Rule again:
$\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Substitute in derivatives:

$\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2} \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}$

Simplify:

$\displaystyle y' = \frac{\big( \tan^2 x + 5x \big) \bigg[ \big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - 2 \sin x \cos x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \bigg]}{\big( \sin^2 x + 6 \big)^2 \big( \csc^2 x + 3 \big)} + \frac{\big( \cos^2 x - 3\sqrt{x} +6 \big) \bigg[ \big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) + 2 \csc^2 x \cot x \big( \tan^2 x + 5x \big) \bigg] }{\big( \csc^2 x + 3 \big)^2 \big( \sin^2 x + 6 \big)}$

We can rewrite the differential by factoring and common mathematical properties to obtain our final answer:

∴ we have found our derivative.

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Learn more about derivatives: brainly.com/question/26836290

Learn more about calculus: brainly.com/question/23558817

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Topic: Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

8 0

2 years ago

Read 2 more answers

Kesha wants to know whether x + 4 is a factor of p(x) = −3x^3 − 15x^2 − 16x −16.

Reika [66]

It's the factor theorem.
P(-4)
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3 0

3 years ago

Jacob can take any one of three routes from school to the mall, and one of five possible routes from the mall to his house. If h

nekit [7.7K]

Answer:

15 routes

Step-by-step explanation:

3 x 5 = 15 routes

8 0

2 years ago

What type of number is v5?

scoundrel [369]

The square root 5 is irrational because a square root of a positive number is either an integer or irrational number. Therefore it is irrational.

7 0

3 years ago

Examine the system of equations.

MrMuchimi

answer:

$x = - 2 \\ y = 2 \\ \\ ( - 2, \: 2)$

explanation:

$y = - \frac{1}{2} x + 1 \\ y = 2x + 6 \\ first \: equation \: add \: \frac{1}{2} from \: both \\ sides \\ \\ second \: equation \: subtract \: 2x \: from \\ both \: sides \\ \\ y + \frac{1}{2} x = 1 \\ y - 2x = 6 \\ solve \: the \: first \: equation \\ \\ y + \frac{1}{2} x = 1 \\ subtract \: \frac{x}{2} from \: both \: sides \\ \\ y = - \frac{1}{2} x + 1 \\ substitute \: - \frac{x}{2} + 1 \: for \: y \: in \: the \\ other \: equation \\ \\ - \frac{1}{2} x + 1 - 2x = 6 \\ add \: \frac{1}{2} x \: to \: 2x \\ \\ - \frac{5}{2} x + 1 = 6 \\ subtract \: 1 \: from \: both \: sides \\ \\ - \frac{5}{2} x = 5 \\ divide \: both \: sides \: by \: - \frac{5}{2} \\ \\ x = - 2 \\ substitute \: the \: value \: of \: x \: into \\ an \: equation \\ \\ y = - \frac{1}{2} ( - 2) + 1 \\ distribute \\ \\ y = 1 + 1 \\ add \\ \\ y = 2 \\ \\ ( - 2, \: 2)$

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3 years ago