# What is the equation of the normal line of #f(x)=sqrt(16x^4-x^3# at #x=4#?

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To find the equation of the normal line to the function f(x) = √(16x^4 - x^3) at x = 4, we need to determine the slope of the tangent line at that point 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, which is f'(x) = (32x^3 - 3x^2) / (2√(16x^4 - x^3)).

Next, we substitute x = 4 into f'(x) to find the slope of the tangent line at x = 4. Evaluating f'(4), we get f'(4) = (32(4)^3 - 3(4)^2) / (2√(16(4)^4 - (4)^3)).

Simplifying this expression, we find f'(4) = 512 / (2√(16384 - 64)).

Further simplifying, we have f'(4) = 512 / (2√(16320)).

Simplifying the denominator, we get f'(4) = 512 / (2 * 128).

This simplifies to f'(4) = 512 / 256 = 2.

Therefore, the slope of the tangent line at x = 4 is 2.

To find the slope of the normal line, we take the negative reciprocal of the tangent line's slope. Thus, the slope of the normal line is -1/2.

Now, we have the slope of the normal line (-1/2) and the point of tangency (4, f(4)). We can use the point-slope form of a line to find the equation of the normal line.

Using the point-slope form, the equation of the normal line is y - f(4) = (-1/2)(x - 4).

Simplifying this equation, we get y - f(4) = (-1/2)x + 2.

Finally, we can rewrite the equation in slope-intercept form by isolating y:

y = (-1/2)x + 2 + f(4).

Therefore, the equation of the normal line to f(x) = √(16x^4 - x^3) at x = 4 is y = (-1/2)x + 2 + f(4).

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