Inverse Functions & Newton's Method: Debmnzl's Question | Yahoo! Answers

[MarkFL]

Here is the question:

A math function question please help?

F(x) = a loge(x) + 1/2
Find the inverse which I found is e^x/a

It then said let a=2 and sketch graph which I did.

I'm getting trouble at this part

Find the value of a in 3 decimal places such that the graphs of y=f(x) and its inverse touch each other once only.

And hence find the contact point.

Please help explanations too please

Here is a link to the question:

A math function question please help? - Yahoo! Answers

I have posted a link there to this topic so that the OP may find my response.

 

[MarkFL]

Hello Debmnzl,

We are given the function:

$\displaystyle f(x)=a\ln(x)+\frac{1}{2}$

and are asked to compute its inverse. So, we may begin by switching $x$ and $f(x)$, and then solving for $f(x)$ which will actually be $\displaystyle f^{-1}(x)$

$\displaystyle x=a\ln(f(x))+\frac{1}{2}$

Subtract through by $\displaystyle \frac{1}{2}$

$\displaystyle x-\frac{1}{2}=a\ln(f(x))$

Get common denominator on the left:

$\displaystyle \frac{2x-1}{2}=a\ln(f(x))$

Divide through by $a$:

$\displaystyle \frac{2x-1}{2a}=\ln(f(x))$

Convert from logarithmic to exponential form:

$\displaystyle e^{\frac{2x-1}{2a}}=f(x)$

and so we have:

$\displaystyle f^{-1}(x)=e^{\frac{2x-1}{2a}}$

Now, as a means of checking our work, we may use the fact that we require:

$\displaystyle f\left(f^{-1}(x) \right)=f^{-1}(f(x))=x$

i) $\displaystyle f\left(f^{-1}(x) \right)=a\ln\left(e^{\frac{2x-1}{2a}} \right)+\frac{1}{2}=a\left(\frac{2x-1}{2a} \right)+\frac{1}{2}=x$

ii) $\displaystyle f^{-1}(f(x))=e^{\frac{2\left(a\ln(x)+\frac{1}{2} \right)-1}{2a}}=e^{\frac{2a\ln(x)}{2a}}=e^{\ln(x)}=x$

So, we know the inverse function we found is correct.

Next, we are instructed to let $a=2$ and sketch the graph. I will assume we are to graph both the given function and its inverse:

View attachment 613

Now, to find a value of $a$ such that the function and its inverse touch only once, we may use the fact that the two are reflected about the line $y=x$, and so must be tangent to this line.

The gradient of the line $y=x$ is 1, and so, we require:

$\displaystyle \frac{d}{dx}\left(f(x) \right)=\frac{a}{x}=1\,\therefore\,a=x$

$\displaystyle \frac{d}{dx}\left(f^{-1}(x) \right)=\frac{e^{\frac{2x-1}{2a}}}{a}=1\,\therefore\,a=e^{\frac{2x-1}{2a}}$

Now since we found $a=x$ we may write:

$\displaystyle a=e^{\frac{2a-1}{2a}}$

To compute the solution to this equation to 3 decimal places, we may use Newton's method. Let's define:

$\displaystyle g(a)=e^{\frac{2a-1}{2a}}-a=0$

where:

$\displaystyle g'(a)=\frac{1}{2a^2}e^{\frac{2a-1}{2a}}-1$

Newton's method gives us the recursion:

$\displaystyle a_{n+1}=a_n-\frac{g(a_n)}{g'(a_n)}$

Using the function we defined and its derivative, we have:

$\displaystyle a_{n+1}=a_n-\frac{e^{\frac{2a_n-1}{2a_n}}-a_n}{\frac{1}{2a_n^2}e^{\frac{2a_n-1}{2a_n}}-1}=\frac{a_n(1-2a_n)}{1-2a^2e^{\frac{1-2a_n}{2a_n}}}$

Using $a_0=2$ as our initial estimate, we then find:

$a_1\approx2.15910252177$

$a_2\approx2.15553677378$

$a_3\approx2.1555352035$

$a_4\approx2.1555352035$

Now, since we were only asked to find the estimate to 3 decimal places, we would write:

$a\approx2.156$

and since we found $a=x$ then the $y$-coordinate of the point of intersection must also be $a$ since the function and its inverse will touch along the line $y=x$

And so the point of intersection is approximately (2.156,2.156). Here is a graph using the better approximation we found:

View attachment 614