Q.Maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ is:
(A) $0$
(B) $12$
(C) $16$
(D) $32$

Concept: Quadratic Extrema — the slope is given by the derivative $y'$, which is a quadratic; its maximum is found at the vertex of that quadratic.

Step 1: Find the slope function.

$y' = -3x^2 + 6x + 9$.

Step 2: This is a downward-opening parabola ($-3 < 0$). Its maximum occurs at

$x = -\frac{b}{2a} = -\frac{6}{2(-3)} = 1$. …

The question asks for the maximum slope of the curve, not the maximum value of the curve itself. That’s a crucial distinction. The slope at any point is given by the derivative $\frac{dy}{dx}$. So we first find the derivative, then find its maximum — because the derivative is itself a function of $x$.

1. Find the slope function. Differentiate $y = -x^3 + 3x^2 + 9x - 27$:

Call this $m(x) = -3x^2 + 6x + 9$. This is a quadratic in $x$, and since the coefficient of $x^2$ is negative ($-3$), its graph is a downward-opening parabola. Such a parabola has a single maximum at its vertex.

2. Find the vertex of $m(x)$. For any quadratic $ax^2 + bx + c$, the vertex occurs at $x = -\frac{b}{2a}$. Here $a = -3$, $b = 6$, so:

So the slope is maximum when $x = 1$.

3. Compute the maximum slope. Substitute $x = 1$ into $m(x)$: $m(1) = -3(1)^2 + 6(1) + 9 = -3 + 6 + 9 = 12.$ …

Method: Optimizing the Slope Function (Second-Layer Extremum)

This method applies whenever a question asks for the maximum or minimum slope of a curve, rather than the maximum or minimum of the curve itself — the slope function has to be optimized as a fresh function in its own right.

Steps

Step 1: Recognize which function is actually being optimized

"Maximum value of $y$" and "maximum slope of $y$" are two different questions. The slope of $y=f(x)$ at any point is the derivative $m(x)=f'(x)$, so "maximum slope" means: treat $m(x)$ itself as a function of $x$ and find ITS maximum — this needs a second round of differentiation, not just $f'(x)=0$.

Step 2: Write down the slope function

Differentiate the given curve once:

Step 3: Optimize the slope function …

Common Mistakes

Mistake 1: Optimizing $y$ instead of the slope function $m(x)=y'$

A student reads "maximum slope" and reflexively sets $\frac{dy}{dx}=0$ to find turning points of the curve itself. That finds where the curve is flat, not where the curve is steepest — a completely different question. The correct move is to differentiate once to get the slope function, and THEN optimize that.

Mistake 2: Forgetting to confirm it is a maximum, not a minimum, of the slope …