# Tangent Line, and Derivative

I was given the function $$f(x)=k\sqrt{x}$$, and a line $$y=x+4$$. I need to find a value for k such that the line is tangent to the graph. I have attempted the problem by taking the derivative of the given function.

## Derivative

$$f'(x)=\frac{k}{2\sqrt{x}}$$

Since the slope of the tangent line is $$1$$, I set the derivative equal to $$1$$ and get:

$$1=\frac{k}{2\sqrt{x}}$$ and then I get: $$2\sqrt{x}=k$$

I feel like I am on the right track, but I am clueless on how to find an x to ensure I find the right k. What other process would be necessary to find k, assuming I am on the right track?

• you are on the right track. the tangent line must also intersect the graph of $f(x)$ in a point, so set $k\sqrt{x}=x+4$ and square both sides to get $k^2 x=x^2+8x+16$. you have already found that $2\sqrt{x}=k$, so $4x=k^2$. now substitute into the first equation and use the quadratic formula – user3816 Oct 19 '16 at 0:11

Hint: You need to use the other information that the $(x,f(x))$ point must lie on the tangent line.
The equation in $x$ for the intersection points of the parabola and the straight line must have a double root.
The equation $\displaystyle{k\,\sqrt{\,x\,}\, = x + 4}$ must have just one solution: $$\left(\,\sqrt{\,x\,}\,\right)^{2} - k\,\sqrt{\,x\,}\, + 4 = 0 \implies \left(\,-k\,\right)^{2} - 4 \times 1 \times 4 = 0\implies k = \pm 4$$ Since $\displaystyle{x + 4 > 0}$ when $\displaystyle{x \geq 0}$, you must take $\bbox[10px,border:1px groove navy]{\displaystyle{k = 4}}$.