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RideAnS [48]
3 years ago
8

F(x)=2x-4. F(x)=2x-4 f-1x=_x+_. Complete the inverse function

Mathematics
1 answer:
natita [175]3 years ago
3 0

Answer:

f^{-1}(x) = \frac{1}{2}(x +4)

Step-by-step explanation:

Given

f(x) = 2x - 4

Required

Determine the inverse function

Start by replacing f(x) with y

y = 2x - 4

Swap the positions of y and x

x = 2y - 4

Make y the subject: <em>Add 4 to both sides</em>

x+4 = 2y - 4+4

x+4 = 2y

Make y the subject: <em>Divide through by 2</em>

<em></em>\frac{1}{2}(x + 4) = y<em></em>

<em></em>y = \frac{1}{2}(x + 4)<em></em>

<em></em>

Replace y with f-1(x)

f^{-1}(x) = \frac{1}{2}(x +4)

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consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where
boyakko [2]

Answer:

a)X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

b)Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

Step-by-step explanation:

From the question we are told that

The Function

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

Generally the differentiation of function f(x) is mathematically solved as

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

f(x)=\frac{x^3+x^2+5x+1}{x^2}

Therefore

f'(x)=\frac{x^2+10x+3}{x^4}

Generally critical point is given as

f'(x)=0

\frac{x^2+10x+3}{x^4}=0

x=-5 \pm\sqrt{22}

Generally the maximum and minimum x value for critical point is mathematically solved as

f'(-5 \pm\sqrt{22})

Where

Maximum value of x

f'(-5 +\sqrt{22})

Minimum value of x

f'(-5 +\sqrt{22})

Therefore interval of increase is mathematically given by

f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})

f(x)

Therefore interval of decrease is mathematically given by

(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)

Generally the second differentiation of function f(x) is mathematically solved as

f''(x)=\frac{2(x^2+15x+6)}{x^5}

Generally the point of inflection is mathematically solved as

f''(x)=0

x^2+15x+6=0

Therefore inflection points is given as

x=\frac{1}{2} (-15 \pm \sqrt{201}

f''(x)>0,\frac{1}{2}(-15-\sqrt{201})

a)Generally the concave upward interval X is mathematically given as

X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

f''(x)

b)Generally the concave downward interval Y is mathematically given as

Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

5 0
3 years ago
I WILL AWARD BRAINLIEST!! PLEASE HELP!! Given: ΔABC, AB = AC X ∈ AC , Y ∈AB AX = AY Prove: BX = CY m∠ABC = m∠ACB △ABX≅△____, by
EleoNora [17]

Answer:

△ABX ≅ △ACY, by reason that they are equal and congruent to each other.

I hope this answers your question.

6 0
3 years ago
Can you shade this please​
Bumek [7]

Answer:

Step-by-step explanation:

✔️It’s a null set

4 0
3 years ago
Steps to solve<br>1/2 / 2/-5 + (-1/4)​
NARA [144]

Answer:

-3/2

Step-by-step explanation:

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7 0
3 years ago
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What is the total price of a $15 meal including 6% sales tax?
aev [14]
15×1.06
=15.9
therefor it is $15.90
4 0
3 years ago
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