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3 years ago

If F(x) and G(x) are two functions such that F(G(x)) = x then F(x) and G(x) are ..?

Mathematics

2 answers:

lana [24]3 years ago

4 0

They are inverse functions though to be completely thorough your teacher should have also put g(f(x)) = x as well. Though I can see what your teacher is aiming for at least.

The idea is that whatever the output of g(x) is, it's plugged into f(x) and the initial input is the result. So g(x) takes a step forward and f(x) takes a step back undoing everything g(x) did. Which is exactly what an inverse operation does.

olga_2 [115]3 years ago

3 0

If f(g(x))=x then they are inverse functions by definition

A is the answer

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kompoz [17]

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4 years ago

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How does 0.03 compare to 0.003

kati45 [8]

0.03 compare to 0.003

While comparing decimals we write the decimals one below the other according to the place value of each digits in place value chart

ones -- point -- Tenths -- Hundredths -- Thousandths

0 . 0 3

0 . 0 0 3

Now we compare each digit left to right. We have 3 in hundredths and 0 in hundredths place. 3 is greater

So 0.03 is greater than 0.003

0.03 > 0.003

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4 years ago

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You have 500 to invest at 3% continuously compounded. How much do you have in five years

tekilochka [14]

The final balance is $580.81.

The total compound interest is $80.81.

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3 years ago

What is the answer to (3-4i)+(2+5i)?

Wewaii [24]

$(3 + 2) + ( - 4i + 5i) \\ 5 + i$

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3 years ago

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Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect

Tasya [4]

$4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1$

defines an ellipse centered at $(0,0)$ with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the $x$ -axis. So if cross sections are taken perpendicular to the $x$ -axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at $x=x_0$ will have a base of length

$\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}$
$\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}$

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a $\sqrt2:1$ ratio with either of the remaining legs. So if the hypotenuse is $\dfrac43\sqrt{9-{x_0}^2}$ , then either leg will have length $\dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}$ .

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is $\dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}$ .

Finally, $x_0$ can be chosen from any value in $-3\le x_0\le3$ . We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

$\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)$

and so the volume would be

$\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx$
$=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}$
$=16$

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3 years ago