All categories

1 year ago

Find the range of the function for the given domain.

Mathematics

1 answer:

Mice21 [21]1 year ago

5 0

Answer: {5, -7, -19, -27, -35}

Step-by-step explanation:

In order solve this, we need to plug in the values of x into the table.

For spaces on the left of the equals sign, you need to write each x from the domain. You can then match that x-value with its function value by putting that on the right side.

For each equation, we are simply plugging a number from the domain into the function and replacing the x-value:

$f(-1) = -4(-1) + 1 = 5\\f(2) = -4(2) + 1 = -7\\f(5) = -4(5) + 1 = -19\\f(7) = -4(7) + 1 = -27\\f(9) = -4(9) + 1 = -35$

I hope this helps. If you need any extra explanation on how the functions are set up, please let me know.

You might be interested in

Martina drove 385 miles in 7 hours. at the same rate, how long would it take her to drive 495 miles?

Anarel [89]

385/7 = 495/x

Answer: 9 hours

5 0

3 years ago

Read 2 more answers

A video store is selling previously owned DVD'S for 40% off the regular price of $15. What is the sale price of the DVD's?

belka [17]

Answer:$9

Step-by-step explanation:

5 0

3 years ago

Which pair of transformations to the figure shown below would produce an image that is on top of the regional

yKpoI14uk [10]

I can’t see any image
Can you please provide the image so i can help

7 0

3 years ago

I am only doing 15 points because I have asked questions and they just take advantage of the points and don't actually answer th

wariber [46]

The answer to your question would be y=35x+50

6 0

3 years ago

I am having trouble with this relative minimum of this equation.<br>

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$