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

Nov 7, 2017

$y = - \frac{2}{3 - e} \left(x - 1\right) + 1 - e$

Explanation:

Let's first find the y-value that corresponds with x = 1 on the function $f \left(x\right) = x \sqrt{x} - {e}^{\sqrt{x}}$:

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

Now, find the derivative of the given function to find an equation for the slope of the tangent line:

$f \left(x\right) = x \sqrt{x} - {e}^{\sqrt{x}}$
$f \left(x\right) = {x}^{\frac{3}{2}} - {e}^{{x}^{\frac{1}{2}}}$ (rewriting f(x) to make it easier to differentiate)

$f p r i m e \left(x\right) = \frac{3}{2} {x}^{\frac{1}{2}} - {e}^{{x}^{\frac{1}{2}}} \cdot \frac{1}{2} {x}^{- \frac{1}{2}}$
$f p r i m e \left(x\right) = \frac{3 \sqrt{x}}{2} - \frac{{e}^{\sqrt{x}}}{2 \sqrt{x}}$

Substitute x = 1 to find the slope of the tangent line at that point:

$f p r i m e \left(1\right) = s l o p e = \frac{3 \sqrt{1}}{2} - \frac{{e}^{\sqrt{1}}}{2 \sqrt{1}}$
$s l o p e = \frac{3 - e}{2}$

The slope of the normal line at a point is the negative reciprocal of the slope of the tangent line at that point. In this case, the slope of the normal line is $- \frac{2}{3 - e}$.

We can now write the equation of the normal line:
$y - y o = m \left(x - x o\right)$
$y - 1 + e = - \frac{2}{3 - e} \left(x - 1\right)$
$y = - \frac{2}{3 - e} \left(x - 1\right) + 1 - e$