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What is the equation of the normal line of #f(x)=xln(3^(1/x))# at #x=9#?

Answer 1

Christopher Agresta

The normal line at #x=9# is the vertical line #x=9#

#f(x) = x[ln(3^(1/x))] = x [1/x ln(3)] = ln(3)#

So the graph of #f# is a horizontal line and the normal line if vertical.

The vertical line with #x=9# has equation #x=9#.

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Answer 2

Emma Aldridge

To find the equation of the normal line of f(x) = xln(3^(1/x)) at x = 9, we need to determine the slope of the tangent line at x = 9 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

First, we find the derivative of f(x) using the product rule and chain rule:

f'(x) = ln(3^(1/x)) + x * (1/x) * (1/(3^(1/x))) * ln(3)

Next, we substitute x = 9 into f'(x) to find the slope of the tangent line at x = 9:

f'(9) = ln(3^(1/9)) + 9 * (1/9) * (1/(3^(1/9))) * ln(3)

Now, we can calculate the slope of the normal line by taking the negative reciprocal of f'(9):

m_normal = -1 / f'(9)

Finally, we can use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line (in this case, (9, f(9))), to find the equation of the normal line.

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