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

Answer 1

Samantha Adkison

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

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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Answer 2

Christopher Agresta

The equation of the line normal to f(x) = xsqrt(x) - e^(sqrt(x)) at x = 1 is y = -2x + 2.

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Answer from HIX Tutor

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