f(x) is cubic polynomial with f(2) = 18 and f(1) = -1. Also f(x) has local maxima at x = -1 and f'(x

Q: Wait, why does $f'(x)$ having a local minimum at $x=0$ mean $f''(0)=0$?

Think of it this way: for any function $g(x)$, a local extremum occurs where its derivative $g'(x)$ is zero. Here, we are looking at the function $g(x) = f'(x)$. Therefore, the extremum of $g(x)$ occurs where $g'(x) = 0$. Since $g'(x)$ is the derivative of $f'(x)$, which is $f''(x)$, it follows that $f''(0) = 0$. It is a beautiful chain of logic!

Q: Why did the $b$ term disappear from the polynomial?

It is not magic, it is geometry! We found that $f''(x) = 6ax + 2b$. When we set $f''(0) = 0$, we get $6a(0) + 2b = 0$, which forces $2b = 0$, meaning $b=0$. This tells us that the inflection point of this specific cubic curve lies exactly on the y-axis. It is a symmetric property of the curve.

Q: How do I know if the function is increasing on the interval $[1, 2\sqrt{5}]$?

A function is strictly increasing when its derivative $f'(x)$ is positive. We found $f'(x) = \frac{57}{4}(x^2 - 1)$. For any value of $x$ greater than $1$, $x^2$ is greater than $1$, making $x^2 - 1$ positive. Since the interval $[1, 2\sqrt{5}]$ is entirely contained within the region where $x > 1$, the derivative remains positive throughout, confirming the function is increasing.

Question

$f(x)$ is cubic polynomial with $f(2) = 18$ and $f(1) = -1$. Also $f(x)$ has local maxima at $x = -1$ and $f'(x)$ has local minima at $x = 0$, then

Options:

(A) the distance between $(-1, 2)$ and $(a, f(a))$, where $x = a$ is the point of local minima is $2\sqrt{5}$

(B) $f(x)$ is increasing for $x \in [1, 2\sqrt{5}]$

(C) $f(x)$ has local minima at $x = 1$

(D) the value of $f(0) = 15$

Sigma Percentile · Accepted Answer

### The Foundation: Defining the Suspect

We begin with the most general form of a cubic polynomial:
$$f(x) = ax^3 + bx^2 + cx + d$$

To understand the behavior of this curve, we need its derivatives. The first derivative, $f'(x) = 3ax^2 + 2bx + c$, tells us about the slope, and the second derivative, $f''(x) = 6ax + 2b$, tells us about the curvature.

### The Inflection Point: The Hidden Clue

The problem drops a fascinating hint: $f'(x)$ has a local minimum at $x = 0$. If the slope function $f'(x)$ has a minimum, its rate of change must be zero at that point.

Mathematically, this means the derivative of the slope—which is $f''(x)$—must be zero at $x = 0$. When we substitute $x = 0$ into $f''(x) = 6ax + 2b$, we get:
$$6a(0) + 2b = 0 \Rightarrow b = 0$$

Suddenly, the term $b$ vanishes! Our polynomial simplifies to:
$$f(x) = ax^3 + cx + d$$

### The Peak and the Valley

Next, we are told the function has a local maximum at $x = -1$. At any local maximum, the tangent line is perfectly horizontal, meaning the slope is zero. Thus, $f'(-1) = 0$.

Substituting $x = -1$ into our simplified derivative $f'(x) = 3ax^2 + c$, we get:
$$3a(-1)^2 + c = 0 \Rightarrow 3a + c = 0 \Rightarrow c = -3a$$

Now, our polynomial is defined by only two variables: $a$ and $d$. We have:
$$f(x) = ax^3 - 3ax + d$$

### The System of Equations

We are given two points: $f(1) = -1$ and $f(2) = 18$. Plugging $x = 1$ into our equation:
$$a(1)^3 - 3a(1) + d = -1 \Rightarrow -2a + d = -1$$

Plugging $x = 2$ into our equation:
$$a(2)^3 - 3a(2) + d = 18 \Rightarrow 8a - 6a + d = 18 \Rightarrow 2a + d = 18$$

We now have a system of linear equations:
$$-2a + d = -1$$
$$2a + d = 18$$

Adding these together, the $a$ terms cancel out, leaving $2d = 17$, so $d = \frac{17}{2}$. Substituting this back, we find $2a = 18 - 8.5 = 9.5$, so $a = \frac{19}{4}$.

### The Grand Reveal

We have successfully reconstructed the polynomial:
$$f(x) = \frac{19}{4}x^3 - \frac{57}{4}x + \frac{17}{2}$$

To find the local minima, we look at the critical points where $f'(x) = 0$. We find $x = 1$ and $x = -1$. Since we know $x = -1$ is the maximum, $x = 1$ must be the minimum.

Finally, we check the monotonicity. Since $f'(x) = \frac{57}{4}(x^2 - 1)$, for any $x > 1$, the derivative is positive, meaning the function is strictly increasing. Thus, the function is **increasing on the interval $[1, 2\sqrt{5}]$**.

Animated Solution for Mathematics - Differentiation: $f (x)$ is cubic polynomial with $f (2) = 18$ and $f (1) = - 1$ . Also $f (x)$ has local maxima at $x = - 1$ and $f^{'} (x)$ has local minima at $x = 0$ , then

Select Answer:

Visualized Solution

Defining the General Cubic Polynomial

Analyzing the Derivative's Minima

Evaluating the Second Derivative

Applying the Local Maxima Condition

Relating Constants $a$ and $c$

Setting up the System of Equations

Solving for Constant $d$

Solving for Constant $a$

Locating the Local Minima

Confirming the Minima Location

Analyzing Function Monotonicity

Final Conclusion

The Sigma Insight: Maxima and Minima

The Foundation

The Inflection Point

The Peak and the Valley

The System of Equations

The Grand Reveal

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