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2 years ago

The equation of a curve is y = x³ + 4x. Show that the value of y increases when x increases. ² is a decreasing function.

Mathematics

1 answer:

belka [17]2 years ago

7 0

We will see that f'(x) > 0, which means that f(x) is an increasing function.

<h3>How to prove that the function is increasing?</h3>

For any function f(x), if f'(x) > 0, then f(x) is increasing for any value of x.

Here we have the cubic function:

f(x) = x³ + 4x

If we differentiate this, we get:

f'(x) = df(x)/dx = 3x² + 4.

And notice that x² is always positive, then f'(x) > 0, which means that f(x) is an increasing function.

If you want to learn more about cubic functions:

brainly.com/question/20896994

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How to solve this trig

n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

First Question

What we have to do is to isolate cos first.

$\displaystyle \large{ cos \theta = - \frac{1}{2} }$

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

Find Q2

$\displaystyle \large{ \pi - \frac{ \pi}{3} = \frac{3 \pi}{3} - \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{2 \pi}{3} }$

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{3} = \frac{3 \pi}{3} + \frac{ \pi}{3} } \\ \displaystyle \large \boxed{ \frac{4 \pi}{3} }$

Both values are apart of the interval. Hence,

$\displaystyle \large \boxed{ \theta = \frac{2 \pi}{3} , \frac{4 \pi}{3} }$

Second Question

Isolate sin(4 theta).

$\displaystyle \large{sin 4 \theta = - \frac{1}{ \sqrt{2} } }$

Rationalize the denominator.

$\displaystyle \large{sin4 \theta = - \frac{ \sqrt{2} }{2} }$

The problem here is 4 beside theta. What we are going to do is to expand the interval.

$\displaystyle \large{0 \leqslant \theta < 2 \pi}$

Multiply whole by 4.

$\displaystyle \large{0 \times 4 \leqslant \theta \times 4 < 2 \pi \times 4} \\ \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}$

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

Find Q3

$\displaystyle \large{ \pi + \frac{ \pi}{4} = \frac{ 4 \pi}{4} + \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{5 \pi}{4} }$

Find Q4

$\displaystyle \large{2 \pi - \frac{ \pi}{4} = \frac{8 \pi}{4} - \frac{ \pi}{4} } \\ \displaystyle \large \boxed{ \frac{7 \pi}{4} }$

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

$\displaystyle \large{ \theta + 2 \pi k = \theta \: \: \: \: \: \sf{(k \: \: is \: \: integer)}}$

Hence:-

For Q3

$\displaystyle \large{ \frac{5 \pi}{4} + 2 \pi = \frac{13 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 4\pi = \frac{21 \pi}{4} } \\ \displaystyle \large{ \frac{5 \pi}{4} + 6\pi = \frac{29 \pi}{4} }$

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

For Q4

$\displaystyle \large{ \frac{ 7 \pi}{4} + 2 \pi = \frac{15 \pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 4 \pi = \frac{23\pi}{4} } \\ \displaystyle \large{ \frac{ 7 \pi}{4} + 6 \pi = \frac{31 \pi}{4} }$

Therefore:-

$\displaystyle \large{4 \theta = \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4} }$

Then we divide all these values by 4.

$\displaystyle \large \boxed{\theta = \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16} }$

Let me know if you have any questions!

3 0

3 years ago

2.06 , 2.6 , 2.9 from smallest to largest

ale4655 [162]

Answer:

2.06 is smallest ans 2.9 is largest

7 0

3 years ago

How do you find the holes, vertical, and horizontal asymptote of an equation?

AfilCa [17]

This is a great question, but it's also a very broad one. Please find and post one or two actual rational expressions, so we can get started on specifics of how to find vertical and horiz. asymptotes.

In the case of vert. asy.: Set the denom. = to 0 and solve for x. Any real x values that result indicate the location(s) of vertical asymptotes.

4 0

3 years ago

HELP! Drag each scenario to show whether the final result will be greater than the original value, less than the original value,

kogti [31]

Answer:

Same as the original:

-->A 15% increase followed by a 15% decrease

--> A 25% decrease followed by a 50% increase

-->A 50% decrease followed by a 100% increase

Greater than the Original:

-->A 20% decrease followed by a 25% decrease

Less than the Original:

-->A 75% increase followed by a 50% decrease

Step-by-step explanation:

Example#1: I did 15% increase is 15% so followed by a -15%(decrease). The sale stayed the same.

Example#2: -20%+25%=5% *greater than the original

Example#3: If the percentage came to 0%, it stayed the same(same as original.)

Hope this helps!!

5 0

3 years ago

Read 2 more answers

Solve for y. 6(y+4)=-3(2y-2) +8y Simplify your answer as much as possible. y =

nadezda [96]

Answer:

6(y + 4) = -3(2y - 2) + 8y

6y + 24 = -6y + 6 + 8y

6y + 24 = 2y + 6

Subtract -2y on both sides

4y + 24 = 6

Subtract -24 on both sides

4y = -18

y = - 4.5

7 0

4 years ago

Read 2 more answers

The equation of a curve is y = x³ + 4x. Show that the value of y increases when x increases. ² is a decreasing function.​

The equation of a curve is y = x³ + 4x. Show that the value of y increases when x increases. ² is a decreasing function.