# What is the equation of the line normal to  f(x)=sqrt(x)-e^(sqrtx) at  x=1?

Feb 6, 2016

$y = \frac{1}{2} \left(1 - e\right) \left(x + 1\right)$

#### Explanation:

We first have to differentiate the function to obtain the gradient of the tangent.

Given $f \left(x\right) = \sqrt{x} - {e}^{\sqrt{x}}$.
Then $f ' \left(x\right) = \frac{1}{2 \sqrt{x}} - \frac{1}{2 \sqrt{x}} {e}^{\sqrt{x}}$.

Now, setting $x = 1$ and substituting into $f ' \left(x\right)$ we can obtain the gradient:

$f ' \left(1\right) = \frac{1}{2 \sqrt{1}} - \frac{1}{2 \sqrt{1}} {e}^{\sqrt{1}} = \frac{1}{2} \left(1 - e\right)$

Now we need a $y$ value for the line to pass through. We know as it is tangent to $f \left(x\right)$ at $x = 1$ then we can substitute $x = 1$ into $f \left(x\right)$ giving us:

$f \left(1\right) = \sqrt{1} - {e}^{\sqrt{1}} = 1 - e$

So we now have an $x$ and $y$ value, all that remains is to substitute the values into $y = m x + c$ to find the y intercept.

$\left(1 - e\right) = \frac{1}{2} \left(1 - e\right) \left(1\right) + c$
$\to c = \frac{1}{2} \left(1 - e\right)$

$y = \frac{1}{2} \left(1 - e\right) x + \frac{1}{2} \left(1 - e\right)$
$y = \frac{1}{2} \left(1 - e\right) \left(x + 1\right)$
Here is a graph showing $f \left(x\right)$ (blue) and the normal line at $x = 1$(orange). (The blue line ends at $x = 0$ due to the trouble of taking the square root of a negative). 