MHB Inverse Functions and "Verifying"

[ardentmed]

Hey guys,

I have a couple more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:

Alright, I'm having quite a bit of trouble with these. So here it goes:

For the first one, I did the 3-step procedure to finding the inverse: write y=f(x), solve for x and y, then interchange variables. Ultimately, this gave me: f^-1(x) = (-x-1)/(3x-2)

And to verify, I just used substitution since f(f^-1(x)) = x:

(2+3)/(6/1) = (2+3)/(6/1) [Is this even remotely correct? I just substituted the corresponding values.]

As for verifying "f^-1 (f(x)) = x, I have no idea how to go about doing this. Do I just substituted the original function in the left hand side into the inverse function?

As for the second question, for 2a, I got t= -a*ln(1-(Q/Qo)) by switching variables to get the inverse and solving for t.

As for question 2b, I assumed that Q/Qo = 0.9 for 90%. Therefore, since a=2, I could substitute that value into the function, giving me:

t= -2ln(1-0.9)
t= 4.6 seconds needed to recharge the battery to 90%.

Any help is much appreciated.

Thanks in advance.

 

[mathmari]

Hey guys,

I have a couple more questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:Alright, I'm having quite a bit of trouble with these. So here it goes:

For the first one, I did the 3-step procedure to finding the inverse: write y=f(x), solve for x and y, then interchange variables. Ultimately, this gave me: f^-1(x) = (-x-1)/(3x-2)

And to verify, I just used substitution since f(f^-1(x)) = x:

(2+3)/(6/1) = (2+3)/(6/1) [Is this even remotely correct? I just substituted the corresponding values.]

As for verifying "f^-1 (f(x)) = x, I have no idea how to go about doing this. Do I just substituted the original function in the left hand side into the inverse function?

As for the second question, for 2a, I got t= -a*ln(1-(Q/Qo)) by switching variables to get the inverse and solving for t.

As for question 2b, I assumed that Q/Qo = 0.9 for 90%. Therefore, since a=2, I could substitute that value into the function, giving me:

t= -2ln(1-0.9)
t= 4.6 seconds needed to recharge the battery to 90%.

Any help is much appreciated.

Thanks in advance.

The inverse function is correct!

To verify that $f(f^{-1}(x)) = x$ :

$$f(f^{-1}(x))=f \left ( \frac{-x-1}{3x-2} \right )=\frac{2 \left (\frac{-x-1}{3x-2} \right )-1}{3\left (\frac{-x-1}{3x-2} \right )+1}=\frac{\frac{-2x-2-3x+2}{3x-2}}{\frac{-3x-3+3x-2}{3x-2}}=\frac{\frac{-5x}{3x-2}}{\frac{-5}{3x-2}}=\frac{-5x}{-5}=x$$

To verify that $f^{-1} (f(x)) = x$ :

$$f^{-1} (f(x))=f^{-1} \left ( \frac{2x-1}{3x+1} \right )=\frac{-\left ( \frac{2x-1}{3x+1} \right )-1}{3\left ( \frac{2x-1}{3x+1} \right )-2}=\frac{\frac{-2x+1-3x-1}{3x+1}}{\frac{6x-3-6x-2}{3x+1}}=\frac{-5x}{-5}=x$$As for the second exercise:

$$Q(t)=Q_0 \left (1-e^{-\frac{t}{a}} \right ) \Rightarrow \frac{Q(t)}{Q_0}=1-e^{-\frac{t}{a}} \Rightarrow e^{-\frac{t}{a}}=1-\frac{Q(t)}{Q_0} \Rightarrow -\frac{t}{a}=\ln{ \left | 1-\frac{Q(t)}{Q_0} \right |} \\ \Rightarrow t=-a \cdot \ln{ \left | 1-\frac{Q(t)}{Q_0} \right |}$$

Your solution for the question $2b$ seems correct to me!