All categories

3 years ago

If f(x) is 4x-3 what is (f^-1of) (10)

2 answers:

6 0

$f(x)=4x-3\ \ \ \ and\ \ \ D_f=R\\\\y=4x-3\ \ \ \Rightarrow\ \ \ y+3=4x\ /:4\ \ \ \Rightarrow\ \ \ \frac{1}{4}y+ \frac{3}{4}=x\\\\f^{-1}(x)= \frac{1}{4}x+ \frac{3}{4} \ \ \ \ and\ \ \ \ D_{f^{-1}}=R\\\\ (f^{-1}\circ f) (10)=f^{-1}\bigg( f (10)\bigg)=\\=f^{-1}\bigg( 4\cdot10-3\bigg)=f^{-1}(37)=\frac{1}{4}\cdot37+ \frac{3}{4} =\frac{37}{4} +\frac{3}{4} = \frac{40}{4} =10$

muminat3 years ago

4 0

$f(x)=4x-3\\
y=4x-3\\
4x=y+3\\
x=\frac{1}{4}y+\frac{3}{4}\\
f^{-1}(x)=\frac{1}{4}x+\frac{3}{4}\\\\
f^{-1} \circ f=\frac{1}{4}(4x-3)+\frac{3}{4}\\
f^{-1} \circ f=\frac{4x}{4}-\frac{3}{4}+\frac{3}{4}\\
f^{-1} \circ f=x$

You might be interested in

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

2 years ago

Please answer!! Will mark Brainliest!!

forsale [732]

Answer:

Is it a multiple choice question

Step-by-step explanation:

8 0

2 years ago

Chapter 10: 10.1-10.4 Quiz > Quiz 10

Bond [772]

1,400 um Idk I’m just really dumb

5 0

3 years ago

a store manager orders t-shirts so that 15 out of 35 are a medium. How many medium shirts would you except to find when there ar

melomori [17]

54 t-shirts 126/35 =3.6

3.6•15=54

6 0

3 years ago

A 2(x + 1) b 2x + 6 c 2x + 2 d 2(x + 3) e 8x

stira [4]

Answer: B and D

Step-by-step explanation:

b. 2x + 6 = 11.4365 and d. 2(x+3) = 11.4365

2/3(9x+6)-1/2(8x-4) = 11.4365

a. 7.4365

c. 7.4365

e. 21.74

8 0

3 years ago