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2 years ago

If f(x) = 3x2 + 1 and g(x) = 1 – x, what is the value of (f – g)(2)?

Mathematics

1 answer:

Blababa [14]2 years ago

6 0

Answer:

$(f-g)(2)=14$

Step-by-step explanation:

We are given the two functions:

$f(x)=3x^2+1\text{ and } g(x)=1-x$

And we want to find the value of:

$(f-g)(2)$

This is equivalent to:

$(f-g)(2)=f(2)-g(2)$

Substitute:

$=(3(2)^2+1)-(1-(2))$

Evaluate:

$=(3(4)+1)-(1-2)=(12+1)-(-1)=13+1=14$

Therefore:

$(f-g)(2)=14$

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Answer:

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Step-by-step explanation:

Firn the lcm of 3, 4 and 6

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1 year ago

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2 years ago

Peter invests $1000 at 4% compounded annually for 10 years and Jose invests $900 compounded annually at 5% for 11 years. Who end

Nesterboy [21]

Answer:

Jose ends up with more money with $59 more than Peter.

Step-by-step explanation:

To determine the amount they will have, you have to use the formula to calculate the future value:

FV=PV(1+r)^n

FV= future value

PV= present value

r= rate of interest

n= number of periods of time

-Peter:

FV=1,000*(1+0.04)^10

FV=1,000*1.48

FV=1,480

-Jose:

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According to this, the answer is that Jose ends up with more money with $59 more than Peter.

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2 years ago

Given f(x) = 4x^4 find f^-1(x) Then state whether f^-1(x) is a function.

sdas [7]

Answer:

$f^{-1}(x)=\pm\sqrt[4]{\dfrac{x}{4}}$
$f^{-1}(x) \quad\text{is not a function}$

Step-by-step explanation:

To find the inverse function, solve for y:

$x=f(y)\\\\x=4y^4\\\\\dfrac{x}{4}=y^4\\\\\pm\sqrt[4]{\dfrac{x}{4}}=y\\\\f^{-1}(x)=\pm\sqrt[4]{\dfrac{x}{4}}$

f(x) is an even function, so f(-x) = f(x). Then the inverse relation is double-valued: for any given y, there can be either of two x-values that will give that result.

___

A function is single-valued. That means any given domain value maps to exactly one range value. The test of this is the "vertical line test." If a vertical line intersects the graph in more than one point, then that x-value maps to more than one y-value.

The horizontal line test is similar. It is used to determine whether a function has an inverse function. If a horizontal line intersects the graph in more than one place, the inverse relation is not a function.

Since the inverse relation for the given f(x) maps every x to two y-values, it is not a function. You can also tell this by the fact that f(x) is an even function, so does not pass the horizontal line test. When f(x) doesn't pass the horizontal line test, f^-1(x) cannot pass the vertical line test.

_____

The attached graph shows the inverse relation (called f₁(x)). It also shows a vertical line intersecting that graph in more than one place.