Problem

(Question 7 continued) In the following diagram, the line $L$ is normal to the graph of $f$ at point $\mathrm{P}$. (d) Find the equation of the line $L$, in the form $y=a x+b$. The line $L$ intersects the graph of $f$ at a second point, $Q$, as shown above. (e) Calculate the distance between $\mathrm{P}$ and $\mathrm{Q}$.

Solution

Step 1 :Find the derivative of the function f(x) and evaluate it at x = 2: \(f'(x) = 3x^2 - 12x + 12\), \(f'(2) = 0\)

Step 2 :Since the slope of the tangent line at point P is 0, the equation of the line L is \(y = 0\)

Step 3 :Solve the equation f(x) = 0: \(x^3 - 6x^2 + 12x - 8 = 0\), x = 2

Step 4 :The second point Q is the same as point P, so the distance between P and Q is \(\boxed{0}\)

From Solvely APP
Source: https://solvelyapp.com/problems/39756/

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