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What is the equation of the normal line of #f(x)= -xln(4^(1-x))# at #x = 1#?

Answer 1

#y = (1-x)/(2ln2)#
or
#y = log_2e/2(1-x)#

#f(x)=-x(1-x)ln4=2x(x-1)ln2# #f^'(x)=2ln2(2x-1)# #f^'(1)=2ln2# #f(1)=0# #m_(|--)=-1/(f^'(1))=-1/(2ln2)=-log_2e/2# #y = -1/(2ln2)(x-1)#

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Answer 2

Emma Aldridge

To find the equation of the normal line of f(x) = -xln(4^(1-x)) at x = 1, we need to determine the slope of the tangent line at x = 1 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) with respect to x:

f'(x) = -ln(4^(1-x)) - x * (1-x) * (ln(4) / 4^(1-x))

Next, we evaluate f'(x) at x = 1 to find the slope of the tangent line:

f'(1) = -ln(4^(1-1)) - 1 * (1-1) * (ln(4) / 4^(1-1)) = -ln(4^0) - 0 * (ln(4) / 4^0) = -ln(1) - 0 * (ln(4) / 1) = 0 - 0 = 0

Since the slope of the tangent line at x = 1 is 0, the slope of the normal line will be the negative reciprocal of 0, which is undefined.

Therefore, the equation of the normal line of f(x) = -xln(4^(1-x)) at x = 1 is undefined.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.