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Georgia [21]
3 years ago
12

What are the points of inflection of k(x)=sinx-(1/4)sin2x

Mathematics
1 answer:
zalisa [80]3 years ago
8 0

Answer:

Option d is correct.

Step-by-step explanation:

The point of inflection of a function y = f(x) at a pointy c is given by f''(c) = 0.

Now, the given function is

k(x) = \sin x - \frac{1}{4}\sin 2x

Differentiating with respect to x on both sides we get,

k'(x) = \cos x - \frac{1}{2} \cos 2x

Again, differentiating with respect to x on both sides we get,

k''(x) = - \sin x + \sin 2x = - \sin x + 2 \sin x \cos x

So, the condition for point of inflection at point c is

k''(c) = 0 = - sin c + 2 sin c cos c

⇒ sin c(2cos c - 1) = 0

⇒ sin c =  0 or \cos c = \frac{1}{2}

⇒ c = 0 or c = \pm \frac{\pi}{3}

Therefore, option d is correct. (Answer)

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4 years ago
Question:
aivan3 [116]

Answer:

part A) The scale factor of the sides (small to large) is 1/2

part B) Te ratio of the areas (small to large) is 1/4

part C) see the explanation

Step-by-step explanation:

Part A) Determine the scale factor of the sides (small to large).

we know that

The dilation is a non rigid transformation that produce similar figures

If two figures are similar, then the ratio of its corresponding sides is proportional

so

Let

z ----> the scale factor

\frac{CB}{C'B'}=\frac{CD}{C'D'}=\frac{BD}{B'D'}

The scale factor is equal to

z=\frac{CB}{C'B'}

substitute

z=\frac{4}{8}

simplify

z=\frac{1}{2}

Part B) What is the ratio of the areas (small to large)?

<em>Area of the small triangle</em>

A=\frac{1}{2}(2)(4)=4\ units^2

<em>Area of the large triangle</em>

A=\frac{1}{2}(4)(8)=16\ units^2

ratio of the areas (small to large)

ratio=\frac{4}{16}=\frac{1}{4}

Part C) Write a generalization about the ratio of the sides and the ratio of the areas of similar figures

In similar figures the ratio of its corresponding sides is proportional and this ratio is called the scale factor

In similar figures the ratio of its areas is equal to the scale factor squared

4 0
4 years ago
Help! What is the average rate of change of f(x)=x^2+3x+6 over the interval -3 less-than-or-equal-to x less-than-or-equal-to 3?
Makovka662 [10]

Answer:

The average rate of change of the function over the interval is 5.

Step-by-step explanation:

Average rate of change of a function:

The average rate of change of a function f(x) over an interval [a,b] is given by:

A = \frac{f(b)-f(a)}{b-a}

Interval -3 less-than-or-equal-to x less-than-or-equal-to 3

This means that a = -3, b = 3

f(x) = x^2 + 3x + 6

So

f(b) = f(3) = (3)^2 + 3(3) + 6 = 9 + 9 + 6 = 24

f(a) = f(-3) = (-3)^2 + 3(-3) + 6 = 9 - 9 + 6 = 6

Average rate of change

A = \frac{f(b)-f(a)}{b-a} = \frac{24+6}{3-(-3)} = \frac{30}{6} = 5

The average rate of change of the function over the interval is 5.

8 0
3 years ago
Help with num 9 please. thanks​
Alik [6]

Answer:

See Below.

Step-by-step explanation:

We want to show that the function:

f(x) = e^x - e^{-x}

Increases for all values of <em>x</em>.

A function is increasing whenever its derivative is positive.

So, find the derivative of our function:

\displaystyle f'(x) = \frac{d}{dx}\left[e^x - e^{-x}\right]

Differentiate:

\displaystyle f'(x) = e^x - (-e^{-x})

Simplify:

f'(x) = e^x+e^{-x}

Since eˣ is always greater than zero and e⁻ˣ is also always greater than zero, f'(x) is always positive. Hence, the original function increases for all values of <em>x.</em>

8 0
3 years ago
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