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3 years ago

F(x) 3x^4-4x^2-3 Is this function odd?

Mathematics

1 answer:

Karo-lina-s [1.5K]3 years ago

8 0

Answer:

No.

Step-by-step explanation:

f(x) = 3x^4-4x^2-3

If a function is odd then f(-x) = -f(x) for all values of x.

Here f(-x) = 3(-x)^4 - 4(-x)^2 - 3

= 3x^4 - 4x^2 - 3

So f(-x) = f(x) which makes it even.

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Given h(x) = -x + 1 find h(0)

g100num [7]

Answer:

Step-by-step explanation:

Just replace 0 instead of x and simplify.

5 0

3 years ago

N Example 3, the points are(0,3500) and (5,1750). Write an equation that represents the distance y (in feet) remaining after x m

Cerrena [4.2K]

Answer:

<h2>The time needed is 10 months.</h2>

Step-by-step explanation:

The given points are (0, 3500) and (5, 1750).

First, we use the formula below to find the slope of the line

$m=\frac{y_{2}-y_{1} }{x_{2} -x_{1} }=\frac{1750-3500}{5-0}=\frac{-1750}{5}=-350$

Which means the function is deacrasing with a ratio of 350 feet per month.

Now, we use the slope and one point to find the equation

$y-y_{1} =m(x-x_{1} )\\y-1750=-350(x-5)\\y=-350x+1750+1750\\y=-350x+3500$

This linear function shows that the situation started at the y-intecept (0, 3500), which means the month 0 had already 3500 feet. In other words, the total distance is 3500 feet. Now, the x-intercept will tell us the time needed to travel that distance.

$0=-350x+3500\\-3500=-350x\\x=\frac{-3500}{-350}\\ x=10$

Therefore, the time needed is 10 months.

3 0

3 years ago

if apples cost$0.59 per pound how would you express the total cost of apples given that you purchase 5 pounds.

tia_tia [17]

Answer:

$2.95

Step-by-step explanation:

0.59*5

2.95

5 0

3 years ago

Read 2 more answers

A model of a building is made using a scale of 1 inch = 25 feet. what is the height of the actual building of the height of the

Solnce55 [7]

1 in. = 25 ft.

12.5 in. = ___ft

25/1=25

12.5*25=312.5

So the answer to your question is 312.5 feet.

Hope this helps! :)

7 0

3 years ago

If x^3 + 1/x^3 = 110 then find x + 1/x

Svetlanka [38]

$Use:(a+b)^3=a^3+3a^2b+3ab^2+b^3\ (*)\\\\x^3+\left(\dfrac{1}{x}\right)^3\\=\underbrace{x^3+3\cdot x^2\cdot\dfrac{1}{x}+3\cdot x\cdot\left(\dfrac{1}{x}\right)^2+\left(\dfrac{1}{x}\right)^3}_{(*)}-3\cdot x^2\cdot\dfrac{1}{x}-3\cdot x\cdot\left(\dfrac{1}{x}\right)^2\\\\=\left(x+\dfrac{1}{x}\right)^3-3x-3\cdot\dfrac{1}{x}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)$

therefore

$x^3+\left(\dfrac{1}{x}\right)^3=110\iff \left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=110\\\\subtitute\ t=x+\dfrac{1}{x}\\\\t^3-3t=110\ \ \ \ |subtract\ 110\ from\ both\ sides\\\\t^3-3t-110=0\\\\t^3-25t+22t-110=0\\\\t(t^2-25)+22(t-5)=0\ \ \ |use\ a^2-b^2=(a-b)(a+b)\\\\t(t-5)(t+5)+22(t-5)=0\\\\(t-5)[t(t+5)+22]=0\iff t-5=0\ or\ \underbrace{t^2+5t+22=0}_{no\ solution}\\\\t=5\iff x+\dfrac{1}{x}=5\\\\Answer:\boxed{x+\dfrac{1}{x}=5}$

3 0

3 years ago