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3 years ago

What is the inverse for the function? y= 2x + 3

2 answers:

7 0

Y = 2x + 3

Let y = f(x), solving for the inverse function is solving for x

y = 2x + 3

2x + 3 = y

2x = y - 3

x = (y - 3)/2

Recall that y = f(x), that implies that x = f⁻¹( y)

x = (y - 3)/2, substitute x = f⁻¹( y)

f⁻¹( y) = (y - 3)/2

if f⁻¹( y) = (y - 3)/2, then: Replace the y with x.

f⁻¹(x) = (x - 3)/2

ivanzaharov [21]3 years ago

4 0

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Can someone PLEASE answer the Algebra Question CORRECTLY BELOW!

Maru [420]

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3 years ago

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!

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3 years ago

Which statement about the function f(x)=|x+3| is true?

Stels [109]

The given function f(x) = |x + 3| has both an absolute maximum and an absolute minimum.

What do you mean by absolute maximum and minimum ?

A function has largest possible value at an absolute maximum point, whereas its lowest possible value can be found at an absolute minimum point.

It is given that function is f(x) = |x + 3|.

We know that to check if function is absolute minimum or absolute maximum by putting the value of modulus either equal to zero or equal to or less than zero and simplify.

So , if we put |x + 3| = 0 , then :

± x + 3 = 0

±x = -3

So , we can have two values of x which are either -3 or 3.

The value 3 will be absolute maximum and -3 will be absolute minimum.

Therefore , the given function f(x) = |x + 3| has both an absolute maximum and an absolute minimum.

Learn more about absolute maximum and minimum here :

brainly.com/question/17438358

#SPJ1

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1 year ago

Solve the equation by finding square roots. To start, remember to isolate x^2. Equation= x^2-9=0

iren2701 [21]

The answer is +3/-3. x^2-9=0 ---> x^2=9 -----> x=+3/-3

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3 years ago

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Likurg_2 [28]

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4 years ago