Problem Analysis #7

To find the value of $c(1) + r(1)^2$, we proceed step by step:

1. Evaluate the function and its derivative at $x = 1$: The given function is:

   $$f(x) = 2x^3 - 4x^2 + 3$$

   At $x = 1$:

   $$f(1) = 2(1)^3 - 4(1)^2 + 3 = 1$$

   The derivative of the function is:

   $$f'(x) = 6x^2 - 8x$$

   At $x = 1$:

   $$f'(1) = 6(1)^2 - 8(1) = -2$$

2. Find the equation of the normal line at $x = 1$: The point of tangency is $P(1, 1)$. The slope of the tangent line at this point is $f'(1) = -2$. The slope of the normal line is the negative reciprocal of the tangent slope:

   $$m_{\text{normal}} = -\frac{1}{f'(1)} = \frac{1}{2}$$

   The equation of the normal line passing through $P(1, 1)$ is:

   $$y - 1 = \frac{1}{2}(x - 1)$$

3. Find the center of the circle $Q(c(1), 0)$: Since the circle is tangent to the curve at $P$, its center $Q(c(1), 0)$ must lie on the normal line. Substituting $y = 0$ into the normal line equation:

   $$0 - 1 = \frac{1}{2}(c(1) - 1)$$ $$-2 = c(1) - 1 \implies c(1) = -1$$

4. Find the radius squared $r(1)^2$: The radius $r(1)$ is the distance between the center $Q(-1, 0)$ and the point $P(1, 1)$:

   $$r(1)^2 = (1 - (-1))^2 + (1 - 0)^2 = 2^2 + 1^2 = 5$$

5. Calculate $c(1) + r(1)^2$:

   $$c(1) + r(1)^2 = -1 + 5 = 4$$

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