# Find k such that the line is tangent to the graph of the function. Function f of x is equal to k...

## Question:

Find {eq}k {/eq} such that the line is tangent to the graph of the function.

Function {eq}f(x)=k\sqrt{x} {/eq}

Line {eq}y=4x+4 {/eq}

{eq}k {/eq}=_____

## Tangent Line:

The tangent line defines the line that best approximates a function at the points near a given point.

For the function {eq}f(x) {/eq} at the point {eq}x=a {/eq}, we can write this tangent line as: {eq}y = f\left( a \right) + f'\left( a \right)\left( {x - a} \right). {/eq}

First, calculating the derivative of the function at a generic point, {eq}x=a {/eq}:

{eq}f(x) = k\sqrt x \quad ,x = a\\ f(a) = k\sqrt a \\ \\ f'(x) = k\frac{1}{{2\sqrt x }} \to f'(a) = k\frac{1}{{2\sqrt a }}\\ \\ y = f\left( a \right) + f'\left( a \right)\left( {x - a} \right)\\ y = k\sqrt a + k\frac{1}{{2\sqrt a }}\left( {x - a} \right)\\ y = k\sqrt a + k\frac{1}{{2\sqrt a }}x - \frac{{\sqrt a }}{2}\\ y = k\frac{1}{{2\sqrt a }}x + k\sqrt a - \frac{{\sqrt a }}{2} {/eq}

Equating the coefficient of this tangent line to the given line {eq}y=4x+4 {/eq}, we have:

{eq}\left\{ \begin{array}{l} k\frac{1}{{2\sqrt a }} = 4\quad \left( 1 \right)\\ k\sqrt a - \frac{{\sqrt a }}{2} = 4\quad \left( 2 \right) \end{array} \right. {/eq}

Combining the equations, we can calculate the values of {eq}k {/eq}:

{eq}\left( 1 \right)\\ k\frac{1}{{2\sqrt a }} = 4 \to \sqrt a = \frac{k}{8}\\ \\ \left( 2 \right)\\ k\sqrt a - \frac{{\sqrt a }}{2} = 4\\ k\frac{k}{8} - \frac{{\frac{k}{8}}}{2} = 4\\ \frac{{{k^2}}}{8} - \frac{k}{{16}} = 4\\ \frac{{2{k^2} - k}}{{16}} = 4\\ 2{k^2} - k = 64\\ 2{k^2} - k - 64 = 0\\ \\ k = \frac{{1 \pm \sqrt {{{\left( { - 1} \right)}^2} - 4 \cdot 2\left( { - 64} \right)} }}{{2 \cdot 2}} = \frac{{1 \pm \sqrt {513} }}{4} {/eq} 