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

1 Answer
Tanini
Nov 7, 2017

y = -2/(3-e) (x - 1) + 1 - e

Explanation:

Let's first find the y-value that corresponds with x = 1 on the function f(x) = xsqrt(x) - e^(sqrt(x)):

f(1) = 1sqrt(1) - e^(sqrt(1))
f(1) = 1 - e

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

f(x) = xsqrt(x) - e^(sqrt(x))
f(x) = x^(3/2) - e^(x^(1/2)) (rewriting f(x) to make it easier to differentiate)

fprime(x) = 3/2x^(1/2) - e^(x^(1/2))*1/2x^(-1/2)
fprime(x) = (3sqrt(x))/2 - (e^sqrt(x))/(2sqrt(x))

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

fprime(1) = slope = (3sqrt(1))/2 - (e^sqrt(1))/(2sqrt(1))
slope = (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 -2/(3-e).

We can now write the equation of the normal line:
y - yo = m(x - xo)
y - 1 + e = -2/(3-e) (x - 1)
y = -2/(3-e) (x - 1) + 1 - e

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