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olga2289 [7]
3 years ago
4

If you vertically stretch the exponential function, f(x) = 2x, by a factor of 4, what is the equation of the new function?

Mathematics
2 answers:
Blababa [14]3 years ago
7 0

Answer: Equation of new function will be f(x)=4(2^x)

Step-by-step explanation:

Since we have given that

The exponential function f(x) is denoted by

f(x)=2^x

We need to stretch the exponential function vertically by a factor of 4.

So, let say c = 4.

So, our new equation becomes,

f(x)=2^x\times c\\\\f(x)=4(2^x)

Hence, equation of new function will be f(x)=4(2^x)

Tema [17]3 years ago
3 0
To vertically strech a function by a factor of c, multiply the whole function by c

so

f(x) vertically streched by a factor of 4 is 4f(x)
so we get
the new function is f(x)=4(2)ˣ
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Answer:

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Step by step explanation:

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Write the Rational Expression in simplest form. <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E2-x-6%7D%7B24-5x-x%5E2%7D" id="T
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-(x+2)/(x+8)

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4 0
4 years ago
A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the number of cakes sold per
HACTEHA [7]

Answer:

A) Revenue function = R(x) = (580x - 10x²)

Marginal Revenue function = (580 - 20x)

B) Fixed Cost = 900

Marginal Cost function = (300 + 50x)

C) Profit function = P(x) = (-35x² + 280x - 900)

D) The quantity that maximizes profit = 4

Step-by-step explanation:

Given,

The Price function for the cake = p = 580 - 10x

where x = number of cakes sold per day.

The total cost function is given as

C = (30 + 5x)² = (900 + 300x + 25x²)

where x = number of cakes sold per day.

Please note that all the calculations and functions obtained are done on a per day basis.

A) Find the revenue and marginal revenue functions [Hint: revenue is price multiplied by quantity i.e. revenue = price × quantity]

Revenue = R(x) = price × quantity = p × x

= (580 - 10x) × x = (580x - 10x²)

Marginal Revenue = (dR/dx)

= (d/dx) (580x - 10x²)

= (580 - 20x)

B) Find the fixed cost and marginal cost function [Hint: fixed cost does not change with quantity produced]

The total cost function is given as

C = (30 + 5x)² = (900 + 300x + 25x²)

The total cost function is a sum of the fixed cost and the variable cost.

The fixed cost is the unchanging part of the total cost function with changing levels of production (quantity produced), which is the term independent of x.

C(x) = 900 + 300x + 25x²

The only term independent of x is 900.

Hence, the fixed cost = 900

Marginal Cost function = (dC/dx)

= (d/dx) (900 + 300x + 25x²)

= (300 + 50x)

C) Find the profit function [Hint: profit is revenue minus total cost]

Profit = Revenue - Total Cost

Revenue = (580x - 10x²)

Total Cost = (900 + 300x + 25x²)

Profit = P(x)

= (580x - 10x²) - (900 + 300x + 25x²)

= 580x - 10x² - 900 - 300x - 25x²

= 280x - 35x² - 900

= (-35x² + 280x - 900)

D) Find the quantity that maximizes profit

To obtain this, we use differentiation analysis to obtain the maximum point of the Profit function.

At maximum point, (dP/dx) = 0 and (d²P/dx²) < 0

P(x) = (-35x² + 280x - 900)

(dP/dx) = -70x + 280 = 0

70x = 280

x = (280/70) = 4

(d²P/dx²) = -70 < 0

Hence, the point obtained truly corresponds to a maximum point of the profit function, P(x).

This quantity demanded obtained, is the quantity demanded that maximises the Profit function.

Hope this Helps!!!

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