What is the equation of the line that is normal to $f (x) = x e^{\sqrt{x}}$ $f(x)= xe^sqrtx$ at $x = 2$ $x= 2$ ?

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Answer 1 · 2016-02-13T01:04:20

$y = e^{\sqrt{2}} (1 + \frac{1}{\sqrt{2}}) x$ $y = e^sqrt(2)(1+1/sqrt(2))x$

To find the equation of the line, we need a point the line passes through and a gradient.

As we wish to find the line that is tangent to $x = 2$ $x=2$ then we can use $f (2)$ $f(2)$ to find a point where the line passes through:

$f (x) = x e^{\sqrt{x}} \to f (2) = 2 e^{\sqrt{2}}$ $f(x)=xe^sqrt(x) -> f(2) = 2e^sqrt(2)$

So we know at least the line passes through $(2, 2 e^{\sqrt{2}})$ $(2, 2e^sqrt(2))$

Now we need the gradient which we can obtain from $f'(x)$ the derivative.

Differentiate $f(x)$ using the product rule and chain rule to get:

$f'(x) = e^sqrt(x)+ x/(2sqrt(x))e^sqrt(x)=e^sqrt(x)(1+sqrt(x)/2)$

We have simplified the function by factoring out $e^sqrt(x)$ and dividing the $x/(2sqrt(x))$ term by $sqrt(x)$ . We can now obtain the gradient:

$f'(2) = e^sqrt(2)(1+sqrt(2)/2)=e^sqrt(2)(1+1/(sqrt(2)))=m$

We can now put this into the equation of a line:

$y-b = m(x-a)$ where $(a,b)$ is our know point from above. That gives us:

$y-2e^sqrt(2) = e^sqrt(2)(1+1/sqrt(2))(x-2)$

We can the $e^sqrt(2)$ over to the right by adding then factoring to simplify and get:

$y = e^sqrt(2)[(1+1/sqrt(2))(x-2)+2]$

What is the equation of the line that is normal to f(x)= xe^sqrtx f(x)=xe√xf(x)= xe^sqrtx at x= 2 x=2 x= 2 ?