Animated Solution for Mathematics - Differentiation: The maximum slope of the curve $y=\frac{1}{2}{x}^4-5x^3+18x^2-19x$ occurs at the point:

Imagine you are a mountaineer trekking along a path defined by the function $y=\frac{1}{2}{x}^4-5x^3+18x^2-19x$. As you walk, the steepness of the ground beneath your feet changes constantly.

In this problem, we are looking for the point where the climb is the most intense—the point of maximum slope.

To find this, we must translate the physical concept of 'steepness' into the language of mathematics. The slope of our path at any point $x$ is given by the first derivative of our position function.

Let us define a new function, $m(x)$, which represents this slope:

Applying the power rule, we find:

This function $m(x)$ is our 'steepness map.' It tells us exactly how steep the path is at any coordinate $x$. Our mission is to find the value of $x$ that makes $m(x)$ as large as possible.

To find the maximum of any function, we look for the 'turning points'—the moments where the function stops increasing and starts decreasing. This occurs when the rate of change of the slope is zero.

We find the derivative of our slope function, $m^{\prime }(x)$, and set it to zero:

Setting $m^{\prime }(x)=0$ gives us a quadratic equation:

Factoring this, we get $(x-2)(x-3)=0$. Our critical points are $x=2$ and $x=3$.

The Final Verdict

We have two candidates, but only one can be the steepest point. To distinguish them, we use the second derivative of the slope, $m^{\prime \prime }(x)$, which describes the concavity of our slope function:

Now, let us test our candidates:

At $x=2$: $m^{\prime \prime }(2)=12(2)-30=-6$. Since this is negative, the slope function is concave down, confirming that $x=2$ is a local maximum.

At $x=3$: $m^{\prime \prime }(3)=12(3)-30=6$. Since this is positive, the slope function is concave up, meaning $x=3$ is a local minimum.

We have identified our $x$-coordinate as $x=2$. To find the final point on the curve, we plug this back into our original position function $y$:

The point $(2,2)$ is where the slope of our curve reaches its maximum value. You have successfully navigated the terrain of calculus, turning a complex curve into a clear, solvable path.