Nested Squareroot

First step , we shall do a little algebra manipulation. We see that: $f\left ( x \right )= \sqrt{x+\sqrt{x+\sqrt{x+...}}}$ $f^{2}\left ( x \right )= x+\sqrt{x+\sqrt{x+...}}$ $f^{2}\left ( x \right )- x = \sqrt{x+\sqrt{x+...}}$ Note that this is same as the initial equation we're manipulating. Thus we may form an equation: $f^{2}\left ( x \right )- x = f\left ( x \right )$ $f^{2}\left ( x \right )-f\left ( x \right )-x=0$ Factorise this equation and we obtain: $\left ( f\left ( x \right )-4 \right )\left ( f\left ( x \right )+3 \right )=0$ Therefore, $f\left ( x \right )=4$ or $f\left ( x \right )=-3$

Second step , now we want to find the equation for the gradient of the function of x. Therefore, we have to find $f'\left ( x \right )$ . We may find this using implicit differentiation on the equation: $f^{2}\left ( x \right )-f\left ( x \right )-x=0$ $2f\left ( x \right )f'\left ( x \right )-f'\left (x \right )-1=0$ $f'\left ( x \right )\left ( 2f\left ( x \right ) -1\right )=1$ $f'\left ( x \right )=\frac{1}{2f\left ( x \right )-1}$

Third step , plugging the value of $f\left ( x \right )=-3$ into $f'\left ( x \right )=\frac{1}{2f\left ( x \right )-1}$ will result in a negative answer, which is contradiction since $a$ and $b$ are positive values. Therefore, we plug in $f\left ( x \right )=4$ into $f'\left ( x \right )=\frac{1}{2f\left ( x \right )-1}$ . Hence: $f'\left ( x \right )=\frac{1}{2\left ( 4 \right )-1}$ $f'\left ( x \right )=\frac{1}{7}$

Our desired answer is $1+7$ = $\boxed{8}$

The gradient of the tangent of the function f is $\nabla f(x) =\frac{\partial f(x)}{\partial x}=\frac{df(x)}{dx}=f'(x),$ so let's work on the function by doing some algebra. First of all write it like this form, $f(x)=\sqrt{x+f(x)},$ derivind using the chain rule $f'(x)=\frac{1+f'(x)}{2\sqrt{x+f(x)}},$ not a lot of algebra we can clear easily f'(x) and get this form $f'(x)=\frac{1}{2\sqrt{x+f(x)}-1},$ and finally doing the substitution on x=12 $f'(12)=\frac{1}{2\sqrt{12+f(12)}-1},$ but we have another problem here; what is f(12) ? So by the definition $f(12)=\sqrt{12+f(12)},$ squaring both sides $f(12)^2=12+f(12) \\ f(12)^2-f(12)-12=0 \\ (f(12)-4)(f(12)+3)=0,$ so taking the positive root we have $f(12)=4,$ and subtituing on the derivate we finally get this $f'(12)=\frac{1}{2\sqrt{12+f(12)}-1} \\ f'(12)=\frac{1}{2\sqrt{12+4}-1}\\ f'(12)=\frac{1}{2\sqrt{16}-1} \\f'(12)=\frac{1}{8-1} \\ f'(12)=\frac{1}{7},$ so $f'(12)=\frac{1}{7}=\frac{a}{b},$ so $a+b=1+7=\boxed{8}$

The question is somewhat confusing I think (Please correct me if I am wrong.)

Question simply wants us to calculate $\frac{df}{dx}$ at $x=12$ .

Now since there is infinite sequence of square roots for $f(x)$ , we can write,

$f(x)=\sqrt{x+f(x)}$

$\therefore f(x)^{2}=x+f(x)$

Differentiating both sides w.r.t.x, we get,

$2f(x)\frac{df}{dx}=1+\frac{df}{dx}$

$\therefore \frac{df}{dx}=\frac{1}{2f(x)-1}$

$\therefore \frac{df}{dx}|_{x=12}=\frac{1}{2f(12)-1}$

Now we want to find out $f(12)$ . Let $f(12)=y$ .

$\therefore y^{2}=12+y$

This gives, $y=4$ and $y=-3$

But the answer which we get using $y=-3$ violets the condition of positiveness asked in the problem hence $y=4$ is the correct choice. This gives:

$\frac{df}{dx}|_{12}=\frac{1}{7}$ and hence, $a+b=\boxed{8}$

first of all a tangent to the function is f'(x) at x=some 'a' (Which is 12 in our case). Our function can be written as $y^{2}$ =x + y. Hence differentiating f(x), we get 2y $\times dy/dx$ = 1 + dy/dx. Now, dy/dx = 1/(2y-1) At x=12, f(x)= $y^{2}$ = 12 + y which gives y=4 or -3. since the gradient of tangent is positive, we have y=4. Plugging which in f'(x) gives us that dy/dx at x=12 is 1/7. Hence the answer 1+7 = $\boxed{8}$

Here is what I did… with some trouble: First, let's get our own definition of the function. $g(x) = \sqrt{x+g(x)}$ So we have a recursive function! Let's find the derivative, so we have a way to find the tangent at $12$ . (I'll use implicit differentiation.) $2g(x)g'(x)=1+g'(x)$ So we have our two functions now. Lets get in the numbers. First, we'll evaluate the original function at 12, and using the quadratic formula, we get: $g(12) = \frac{1+\sqrt{1+48}}{2} = 4$ Then back to the derivative… $g'(12) = \frac{1}{2g(12)-1}=\frac{1}{8-1}$ Finally: $\boxed{a+b = 8}$

Let $N$ be the RHS, and let $y = f(x)$ , Then, we have

$y = N$

$y^2 = N + x$

Thus, we have $y = y^2 - x$ and $y^2 - y = x$ . Taking the implicit derivative, we have $2y \frac{dy}{dx} - \frac{dy}{dx} = 1$ and $\frac{dy}{dx} = \frac{1}{2y - 1}$ . We require to find the value of $y = N$ at $x = 12$ .

So, going back to our equation, we have that $y^2 - y = 12 \rightarrow (y - 4)(y + 3) = 0$ . $y$ has to be a positive value, so $y = 4$ . Our derivative of the tangent line is then $\frac{dy}{dx} = \frac{1}{7}$ .

Let's find an explicit solution for $f(x)$ . As others have noted, it must satisfy: $f(x) = \sqrt{x + f(x)}$ which, letting $y = f(x)$ , we can rearrange as: $y^2 - y = x$ We recognise this as a parabola that has been reflected in the line $y = x$ and has y-intercepts at $y = 0, 1$ . The quadratic formula gives us two possible solutions, $y = \frac{1}{2} \pm \frac{1}{2}\sqrt{1 + 4x}$ which are the upper and lower branches of the parabola. Since the tangent to the desired function is positive at $x = 12$ , we are only interested in the upper branch. The derivative of it is: $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\sqrt{1 + 4x}}$ Evaluating at $x = 12$ we get $f^\prime(12) = \frac{1}{7}$ and so the answer is $1 + 7 = \boxed{8}$ .

It is a great problem love it!

Let us consider y = f(x)

Then y^{2} = x + y

Differentiating with respect to x,

         [2 \times y \times \frac{dy}{dx}] = [1 + \frac{dy}{dx}]

          Gradient      [ \frac{dy}{dx}] = [\frac{1}{ 2 \times y - 1}]


          when x =12, y= 4 or (-3)

          Since only positive value is asked, y = 4.

          so,    y =4, [\frac{dy}{dx} ]= [\frac{1{7}] = [\frac{a}{b}]

          so, [\boxed{a + b = 8}]

y^2=x+y or, y^2-y-12=0 or, y=-3,4. As y can be expressed as the square root of real number,the value of y can't be negetive. So, y=4.

Now,(dy/dx)=1/(2y-1) for y=4,the value of (dy/dx)=1/7=a/b

So, a+b=8

Let $y = f(x)$ .

Find the solution of $f(12)$

As soon as $y = \sqrt{x+\sqrt{x+...}}$

$y= \sqrt{x+y}$

$y^2-y=x$

At $x=12, \boxed{y=4}$ .

Deriving f(x)

$\frac{d y^2-y}{d x} = \frac{d x}{d x}$

• By the chain rule:

$\frac{d y^2-y}{d y} y' = 1$

$\boxed{y' = \frac{1}{2y-1}}$

Concluding

$y'= \frac{a}{b}$

$y' = \frac{1}{7}$

$\boxed{a+b=8}$

It forms a quadratic equation in f(x) , since the pattern inside the root is infinite...

f(x)= sqrt(x + f(x)) , which means, after solving using Vieta's formula,

2*f(x) = 1+ sqrt(4x+1)
{after ignoring the other solution having '-' sign with sqrt(4x+1), since f(x) would always be positive}

Differentiating this value of f(x) w.r.t 'x', we get,

f'(x) = 1/sqrt(4x+1)

which gives f'(12) = 1/7 ...