If f(x) = x³ + bx² + cx + d and 0

Q: Wait, why do we reject the possibility of local maxima or minima?

That is a fantastic question! A local maximum or minimum occurs only where the derivative $f'(x)$ changes sign, which requires the derivative to be zero at some point. Because our discriminant $D$ is strictly negative, the derivative $f'(x)$ never touches the x-axis—it is always positive. Since the slope is always positive, the function never 'turns around' to create a peak or a valley. It is a monotonic climb!

Q: Why can't we just use energy conservation or some other physics principle here?

I love the interdisciplinary thinking! While energy conservation is a powerful tool in physics, this is a pure calculus problem. We are analyzing the geometric properties of a cubic polynomial. The 'energy' here is the function's value, and the 'force' is its slope. By ensuring the slope is always positive, we guarantee the function is strictly increasing, much like a particle moving in one direction without ever stopping.

Q: Is it possible for $f(x)$ to be bounded if it is strictly increasing?

Think about the end behavior of a cubic function. As $x \to \infty$, $f(x) \to \infty$, and as $x \to -\infty$, $f(x) \to -\infty$. A strictly increasing function that spans from negative infinity to positive infinity cannot be bounded. It must traverse the entire vertical range!

Question

If $f(x) = x^3 + bx^2 + cx + d$ and $0 < b^2 < c$, then in $(-\infty, \infty)$

Options:

(A) $f(x)$ is a strictly increasing function

(B) $f(x)$ has a local maxima

(C) $f(x)$ is a strictly decreasing function

(D) $f(x)$ is bounded

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, infinite plain, and before you lies the graph of a cubic function:
$$f(x) = x^3 + bx^2 + cx + d$$

This looks like a smooth, winding road stretching from the depths of negative infinity to the heights of positive infinity. Today, we have a special constraint: $0 < b^2 < c$. This inequality is the key that unlocks the secret behavior of this curve.

### The First Step: Finding the Slope

To understand the motion of our curve, we need to know its slope at every point. In calculus, the slope is defined by the first derivative, $f'(x)$.

Using the power rule, where the derivative of $x^n$ is $nx^{n-1}$, we differentiate our function term by term:
$$f'(x) = \frac{d}{dx}(x^3 + bx^2 + cx + d)$$

This yields the quadratic expression:
$$f'(x) = 3x^2 + 2bx + c$$

Notice the transformation! We started with a cubic function and arrived at a quadratic expression. This quadratic is the "engine" that drives the behavior of our original function.

### The Geometry of the Derivative

Consider the expression $f'(x) = 3x^2 + 2bx + c$. This represents a parabola. Since the leading coefficient is $3$ (which is positive), the parabola is a "smiling" face opening upwards.

To determine if it ever crosses the $x$-axis, we examine the discriminant, $D = B^2 - 4AC$. Here, $A=3$, $B=2b$, and $C=c$.

Substituting these values, we get:
$$D = (2b)^2 - 4(3)(c) = 4b^2 - 12c$$

We can factor this as:
$$D = 4(b^2 - 3c)$$

### The Magic of the Inequality

Now, let's apply our given condition: $0 < b^2 < c$. Since $c$ is positive, it follows that $c < 3c$.

Because $b^2 < c$, it must be true that $b^2 < 3c$. If we subtract $3c$ from both sides, we get:
$$b^2 - 3c < 0$$

This is the breakthrough! Our discriminant $D = 4(b^2 - 3c)$ is strictly negative. A negative discriminant means the parabola $f'(x)$ has no real roots and never touches the $x$-axis.

Since it opens upwards and never touches the axis, it must be floating entirely above the $x$-axis. Mathematically, this means:
$$f'(x) > 0 \quad 	ext{for all } x \in (-\infty, \infty)$$

### The Final Victory

If the derivative is always positive, the slope is always positive. If the slope is always positive, the function $f(x)$ is **strictly increasing**.

There are no peaks, no valleys, and no flat spots. It is a relentless, beautiful ascent. You have just proven that under the condition $0 < b^2 < c$, the cubic function is a **strictly increasing function**.

Animated Solution for Mathematics - Differentiation: If $f (x) = x^{3} + b x^{2} + c x + d$ and $0 < b^{2} < c$ , then in $(- \infty, \infty)$

Select Answer:

Visualized Solution

Understand the Function $f (x)$

The Tool for Monotonicity: $f^{'} (x)$

Calculate the Derivative $f^{'} (x)$

Analyzing the Quadratic $f^{'} (x)$

Calculate the Discriminant $D$

Evaluate the Sign of $D$

Conclusion: Strictly Increasing

The Sigma Insight: Monotonicity

Analyzing the Setup

The First Step

The Geometry of the Derivative

The Magic of the Inequality

The Final Victory

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=x3+bx2+cx+d and 0<b2<c, then in (−∞,∞)

Select Answer:

Visualized Solution

Understand the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Tool for Monotonicity: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Calculate the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Analyzing the Quadratic .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Calculate the Discriminant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Evaluate the Sign of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D