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Animated Solution for Mathematics - Differentiation: If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1, a > 0$ has a local maximum at $x = α$ and a local minimum at $x = α^{2}$ , then $α$ and $α^{2}$ are the roots of the equation :

Q: Wait, why do we reject the negative root when solving $a(a - 2) = 0$?

That is a fantastic catch! In algebra, $a(a - 2) = 0$ gives us $a = 0$ or $a = 2$. However, look closely at the problem statement: it explicitly states $a > 0$. If $a$ were $0$, the function would become $f(x) = 2x^3 + 1$, which is a strictly increasing function with no local maxima or minima. Always check your constraints—they are the boundaries of your mathematical reality!

Q: Why can't we just use the first derivative to find the max and min?

The first derivative $f'(x) = 0$ only tells us where the slope is zero—it identifies the 'candidates' for extrema. It doesn't tell us if that point is a peak or a valley. To distinguish between them, we must use the second derivative test (concavity) or analyze the sign change of the first derivative around the critical point.

Q: Is there a faster way to find the quadratic equation without solving for $a$?

Yes! Once you know the roots are $2$ and $4$, you can use Vieta's formulas. For a quadratic $x^2 - (sum)x + (product) = 0$, the sum of roots is $2 + 4 = 6$ and the product is $2 \times 4 = 8$. This gives you $x^2 - 6x + 8 = 0$ immediately. It is a great shortcut for competitive exams!

Select Answer:

Progressive Hints

Hint 1

Visualize the cubic curve. When you are standing at the very top of a hill (local maximum) or the bottom of a valley (local minimum), what is the slope of the tangent line? Remember, the derivative

f^{'} (x)

represents the slope of the tangent. If the slope is zero, you have found a critical point.

Hint 2

You have two critical points, but how do you know which is the peak and which is the valley? You need the Second Derivative Test. Calculate

f^{''} (x)

. If

f^{''} (x) < 0

, the curve is concave down (a hill/maximum). If

f^{''} (x) > 0

, it is concave up (a valley/minimum).

Hint 3

You have identified the critical points as

x = a

and

x = 2 a

. The problem gives you the locations as

α

and

α^{2}

. Set up the system:

α = a

and

α^{2} = 2 a

. Now, solve for

a

by substituting the first into the second.

Visualized Solution

The Cubic Curve and its Extrema

Given function: $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$
Constraint: $a > 0$
Local maximum occurs at $x = α$
Local minimum occurs at $x = α^{2}$

Condition for Local Extrema

To find local maxima and minima, we need the critical points.
Critical points occur where the first derivative is zero.
Set $f^{'} (x) = 0$

Differentiating $f (x)$

$f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1$
Differentiating with respect to $x$ :
$f^{'} (x) = \frac{d}{d x} (2 x^{3}) - \frac{d}{d x} (9 a x^{2}) + \frac{d}{d x} (12 a^{2} x) + \frac{d}{d x} (1)$
$f^{'} (x) = 6 x^{2} - 18 a x + 12 a^{2}$

Factoring the Derivative

$f^{'} (x) = 6 (x^{2} - 3 a x + 2 a^{2})$
Splitting the middle term: $- 3 a x = - 2 a x - a x$
$f^{'} (x) = 6 (x^{2} - 2 a x - a x + 2 a^{2})$
$f^{'} (x) = 6 (x - a) (x - 2 a)$

Identifying the Critical Points

Set $f^{'} (x) = 0$ to find critical points.
$6 (x - a) (x - 2 a) = 0$
The critical points are $x = a$ and $x = 2 a$ .
Since $a > 0$ , we know that $a < 2 a$ .

Second Derivative Test

Find the second derivative $f^{''} (x)$ to check concavity.
$f^{''} (x) = \frac{d}{d x} (6 x^{2} - 18 a x + 12 a^{2})$
$f^{''} (x) = 12 x - 18 a$

Evaluating Extrema at $x = a$

Substitute $x = a$ into $f^{''} (x)$ :
$f^{''} (a) = 12 (a) - 18 a = - 6 a$
Since $a > 0$ , $- 6 a < 0$ .
A negative second derivative means $x = a$ is a point of local maximum.

Evaluating Extrema at $x = 2 a$

Substitute $x = 2 a$ into $f^{''} (x)$ :
$f^{''} (2 a) = 12 (2 a) - 18 a = 24 a - 18 a = 6 a$
Since $a > 0$ , $6 a > 0$ .
A positive second derivative means $x = 2 a$ is a point of local minimum.

Equating to Given Roots

The problem states the local maximum is at $x = α$ .
We found the local maximum is at $x = a$ .
Therefore, $α = a$ .
The problem states the local minimum is at $x = α^{2}$ .
We found the local minimum is at $x = 2 a$ .
Therefore, $α^{2} = 2 a$ .

Solving for $a$

We have the system: $α = a$ and $α^{2} = 2 a$ .
Substitute $α = a$ into the second equation:
$a^{2} = 2 a$
$a^{2} - 2 a = 0$
$a (a - 2) = 0$
Since $a > 0$ , we reject $a = 0$ . Thus, $a = 2$ .

Calculating the Roots

Now that we know $a = 2$ , we can find the roots.
First root: $α = a = 2$
Second root: $α^{2} = 2^{2} = 4$
The roots of our required quadratic equation are $2$ and $4$ .

Forming the Final Quadratic Equation

A quadratic equation with roots $r_{1}$ and $r_{2}$ is $(x - r_{1}) (x - r_{2}) = 0$ .
Substitute the roots $2$ and $4$ :
$(x - 2) (x - 4) = 0$
Expanding the brackets: $x^{2} - 4 x - 2 x + 8 = 0$
Final Equation: $x^{2} - 6 x + 8 = 0$

00:00 / 00:00

The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

Imagine you are walking along the path of a roller coaster defined by the function

f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1

. As you traverse this path, you encounter a peak—a local maximum—and a valley—a local minimum.

In the language of calculus, these points are the moments where the slope of your path is perfectly horizontal. This is the fundamental insight that unlocks the entire problem.

The Calculus Engine

To find these stationary points, we must invoke the power of the derivative. By setting

f^{'} (x) = 0

, we identify the locations where the function is "flat."

Let us differentiate

f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1

term by term:

f^{'} (x) = \frac{d}{d x} (2 x^{3}) - \frac{d}{d x} (9 a x^{2}) + \frac{d}{d x} (12 a^{2} x) + \frac{d}{d x} (1)

This yields the quadratic derivative:

f^{'} (x) = 6 x^{2} - 18 a x + 12 a^{2}

To simplify, we factor out the common

6

f^{'} (x) = 6 (x^{2} - 3 a x + 2 a^{2})

By splitting the middle term, we find the roots of this derivative are

x = a

and

x = 2 a

. These are our critical points.

The Concavity Test

We must use the Second Derivative Test to determine the nature of these points. Let us differentiate again:

f^{''} (x) = 12 x - 18 a

Now, we test our critical points. For

x = a

, we get

f^{''} (a) = 12 (a) - 18 a = - 6 a

. Since the problem guarantees

a > 0

, this value is negative, confirming that

x = a

is our local maximum.

Conversely, for

x = 2 a

, we get

f^{''} (2 a) = 12 (2 a) - 18 a = 6 a

. This is positive, confirming that

x = 2 a

is our local minimum.

The Synthesis

The problem states the local maximum is at

x = α

and the local minimum is at

x = α^{2}

. Matching our findings, we establish the system:

α = a

α^{2} = 2 a

Substituting the first into the second, we get

α^{2} = 2 α

, which simplifies to

α (α - 2) = 0

. Since

a > 0

, we discard

α = 0

and find

α = 2

Consequently, our roots are

α = 2

and

α^{2} = 4

Final Calculation

Finally, we construct the quadratic equation using these roots. A quadratic with roots

r_{1}

and

r_{2}

is given by

(x - r_{1}) (x - r_{2}) = 0

. Substituting

2

and

4

(x - 2) (x - 4) = 0

Expanding this, we arrive at the elegant solution:

$x^{2} - 6 x + 8 = 0$

Common Traps & FAQs

Wait, why do we reject the negative root when solving

a (a - 2) = 0

That is a fantastic catch! In algebra,

a (a - 2) = 0

gives us

a = 0

a = 2

. However, look closely at the problem statement: it explicitly states

a > 0

. If

a

were

0

, the function would become

f (x) = 2 x^{3} + 1

, which is a strictly increasing function with no local maxima or minima. Always check your constraints—they are the boundaries of your mathematical reality!

Why can't we just use the first derivative to find the max and min?

The first derivative

f^{'} (x) = 0

only tells us where the slope is zero—it identifies the 'candidates' for extrema. It doesn't tell us if that point is a peak or a valley. To distinguish between them, we must use the second derivative test (concavity) or analyze the sign change of the first derivative around the critical point.

Is there a faster way to find the quadratic equation without solving for

a

Yes! Once you know the roots are

2

and

4

, you can use Vieta's formulas. For a quadratic

x^{2} - (s u m) x + (p r o d u c t) = 0

, the sum of roots is

2 + 4 = 6

and the product is

2 \times 4 = 8

. This gives you

x^{2} - 6 x + 8 = 0

immediately. It is a great shortcut for competitive exams!

Animated Solution for Mathematics - Differentiation: If the function $f (x) = 2 x^{3} - 9 a x^{2} + 12 a^{2} x + 1, a > 0$ has a local maximum at $x = α$ and a local minimum at $x = α^{2}$ , then $α$ and $α^{2}$ are the roots of the equation :

Select Answer:

Visualized Solution

The Cubic Curve and its Extrema

Condition for Local Extrema

Differentiating $f (x)$

Factoring the Derivative

Identifying the Critical Points

Second Derivative Test

Evaluating Extrema at $x = a$

Evaluating Extrema at $x = 2 a$

Equating to Given Roots

Solving for $a$

Calculating the Roots

Forming the Final Quadratic Equation

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Calculus Engine

The Concavity Test

The Synthesis

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

The Cubic Curve and its Extrema

Condition for Local Extrema

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

Factoring the Derivative

Identifying the Critical Points

Second Derivative Test

Evaluating Extrema at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=a

Evaluating Extrema at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2a

Equating to Given Roots

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Calculating the Roots

Forming the Final Quadratic Equation

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Calculus Engine

The Concavity Test

The Synthesis

Final Calculation

Similar Questions

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