Key Concepts and Formulas

• Slope of a Curve: The slope of a curve $y = f(x)$ is given by its first derivative, $\frac{dy}{dx} = f'(x)$.
• Maxima and Minima: To find the maxima or minima of a function $f(x)$, find the critical points by setting $f'(x) = 0$. Then, use the second derivative test: if $f''(x) > 0$, it's a local minimum; if $f''(x) < 0$, it's a local maximum.
• Second Derivative Test applied to Slope: To find the maximum slope, we need to maximize the slope function $M(x) = \frac{dy}{dx}$. We find where $\frac{dM}{dx} = \frac{d^2y}{dx^2} = 0$, and then check the sign of $\frac{d^2M}{dx^2} = \frac{d^3y}{dx^3}$ at these critical points.

Step-by-Step Solution

Step 1: Define the Original Curve and Calculate its Slope Function

We are given the curve: $y = \frac{1}{2}x^4 - 5x^3 + 18x^2 - 19x$

To find the slope of this curve, we differentiate $y$ with respect to $x$: $M(x) = \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2}x^4 - 5x^3 + 18x^2 - 19x\right)$ $M(x) = \frac{1}{2}(4x^3) - 5(3x^2) + 18(2x) - 19$ $M(x) = 2x^3 - 15x^2 + 36x - 19$

The function $M(x)$ represents the slope of the curve at any point $x$.

Step 2: Find Critical Points of the Slope Function

To find the maximum slope, we need to find the critical points of $M(x)$. We differentiate $M(x)$ with respect to $x$ and set it equal to zero. $\frac{dM}{dx} = \frac{d}{dx}\left(2x^3 - 15x^2 + 36x - 19\right)$ $\frac{dM}{dx} = 6x^2 - 30x + 36$

Now, set $\frac{dM}{dx} = 0$ to find the critical points: $6x^2 - 30x + 36 = 0$ Divide by 6: $x^2 - 5x + 6 = 0$ Factor the quadratic: $(x - 2)(x - 3) = 0$ So, the critical points are $x = 2$ and $x = 3$.

Step 3: Apply the Second Derivative Test to the Slope Function

To determine whether the critical points correspond to a maximum or minimum slope, we need to find the second derivative of $M(x)$, which is the third derivative of the original function $y$.

$\frac{d^2M}{dx^2} = \frac{d}{dx}\left(6x^2 - 30x + 36\right)$ $\frac{d^2M}{dx^2} = 12x - 30$

Now, we evaluate $\frac{d^2M}{dx^2}$ at the critical points $x = 2$ and $x = 3$:

• For $x = 2$: $\frac{d^2M}{dx^2}(2) = 12(2) - 30 = 24 - 30 = -6$ Since $\frac{d^2M}{dx^2}(2) < 0$, the slope is maximum at $x = 2$.

• For $x = 3$: $\frac{d^2M}{dx^2}(3) = 12(3) - 30 = 36 - 30 = 6$ Since $\frac{d^2M}{dx^2}(3) > 0$, the slope is minimum at $x = 3$.

We are looking for the maximum slope, so we choose $x = 3$.

Step 4: Find the Corresponding y-coordinate

Now we substitute $x = 3$ into the original equation to find the corresponding y-coordinate: $y = \frac{1}{2}(3)^4 - 5(3)^3 + 18(3)^2 - 19(3)$ $y = \frac{1}{2}(81) - 5(27) + 18(9) - 57$ $y = \frac{81}{2} - 135 + 162 - 57$ $y = \frac{81}{2} - 30$ $y = \frac{81 - 60}{2} = \frac{21}{2}$

So, the point where the slope is maximum is $\left(3, \frac{21}{2}\right)$.

Common Mistakes & Tips

• Confusing Maximum Slope with Maximum/Minimum of the Curve: The problem asks for the maximum slope, not the maximum or minimum y-value of the original curve.
• Forgetting to find the y-coordinate: Once you find the x-coordinate where the slope is maximum, remember to plug it back into the original equation to find the corresponding y-coordinate.
• Sign Errors: Pay close attention to signs when differentiating and evaluating expressions.

Summary

To find the maximum slope of the given curve, we first found the slope function by taking the first derivative. Then, we found the critical points of the slope function by setting its derivative (the second derivative of the original function) equal to zero. We used the second derivative test on the slope function (the third derivative of the original function) to determine which critical point corresponded to a maximum slope. Finally, we substituted the x-coordinate of the maximum slope back into the original equation to find the corresponding y-coordinate. The point where the slope is maximum is $\left(3, \frac{21}{2}\right)$.

Final Answer

The final answer is $\boxed{\left( {3,{{21} \over 2}} \right)}$, which corresponds to option (A).