What is the equation of the line that is normal to f(x)=4x3−3x2+5x−2 at x=1?

Question

cutimnm135imsa · Accepted Answer

The slope of the normal at a given point is the negative inverse of the slope of the function at that point. i.e. if the slope of the function is m then the slope of the normal is

- \frac{1}{m}

To find the slope, differentiate the function.

f^{'} (x) = 12 x^{2} - 6 x + 5

At x=1 this equals 12-6+5=11
Therefore the slope of the normal is

- \frac{1}{11}

The calculate the constant c in the standard lien equation y=mx+c, substitute
x=1 back into the original function f(x)

= 4 \cdot 1^{3} - 3 \cdot 1^{2} + 5 \cdot 1 - 2 = 4

Therefore the normal is

y = (- \frac{1}{11}) x + 4

11y=-x+44

ruseducatives1t03 · Accepted Answer

Find the value of the function at the given point:

y_{0} = f (1) = 4

.
The slope of the normal line at

x = x_{0}

is the negative reciprocal of the derivative of the function, evaluated at

x = x_{0} : M (x_{0}) = - \frac{1}{f^{'} (x_{0})}

Find the derivative:

f^{'} (x) = {(4 x^{3} - 3 x^{2} + 5 x - 2)}^{'} = 12 x^{2} - 6 x + 5

Hence,

M (x_{0}) = - \frac{1}{f^{'} (x_{0})} = - \frac{1}{12 x_{0}^{2} - 6 x_{0} + 5}

Next, find the slope at the given point.

m = M (1) = - \frac{1}{11}

Finally, the equation of the normal line is

y - y_{0} = m (x - x_{0})

Plugging the found values, we get that

y - 4 = - \frac{x - 1}{11}

.
Or, more simply:

y = \frac{45}{11} - \frac{x}{11}

.

What is the equation of the line that

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