What is the equation of the line normal to $f (x) = x^{3} - 3 x^{2}$ $f(x)=x ^3-3x^2$ at $x = 4$ $x=4$ ?

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Answer 1 · 2017-08-01T19:02:37

$y = \frac{1}{24} x - \frac{23}{6}$ $y = frac(1)(24) x - frac(23)(6)$

We have: $f (x) = x^{3} - 3 x^{2}$ $f(x) = x^(3) - 3 x^(2)$

First, let's find the $y$ $y$ -coordinate corresponding to the given value of $x$ $x$ :

$\Rightarrow f (4) = {(4)}^{3} - 3 \cdot {(4)}^{2}$ $Rightarrow f(4) = (4)^(3) - 3 cdot (4)^(2)$

$\Rightarrow f (4) = 64 - 48$ $Rightarrow f(4) = 64 - 48$

$therefore f(4) = 16$

So we must find the equation of the normal line at the point $(4, 16)$ .

Then, let's differentiate $f(x)$ :

$Rightarrow f'(x) = 3 x^(2) - 6 x$

For $x = 4$ , the slope of the line is:

$Rightarrow f'(4) = 3 cdot (4)^(2) - 6 cdot (4)$

$Rightarrow f'(4) = 48 - 24$

$therefore f'(4) = 24$

The gradient of the normal line is the reciprocal of this value:

$Rightarrow "Gradient of normal line" = (f'(4))^(- 1)$

$therefore "Gradient of normal line" = frac(1)(24)$

Now, let's express the equation of the normal line in point-slope form:

$Rightarrow y - y_(1) = m (x - x_(1))$

$Rightarrow y - 16 = frac(1)(24) (x - 4)$

$Rightarrow y - 4 = frac(1)(24) x - frac(1)(6)$

$therefore y = frac(1)(24) x - frac(23)(6)$

Therefore, the equation of the normal line is $y = frac(1)(24) x - frac(23)(6)$ .

What is the equation of the line normal to f(x)=x ^3-3x^2 f(x)=x3−3x2f(x)=x ^3-3x^2 at x=4x=4x=4?