For a twice differentiable function f(x), g(x) is defined as g(x) = (f'(x))² + f''(x)f(x) on [a, e].

Q: Wait, why can't we just use the Mean Value Theorem directly on $g(x)$?

The Mean Value Theorem is powerful, but it usually tells us about the existence of a derivative value at a single point. Here, we are looking for the *number* of zeros of a derivative, which is the classic domain of Rolle's Theorem. By identifying $g(x)$ as $h'(x)$, we transform a complex expression into a simple derivative problem.

Q: How do I know for sure that the zeros of $f(x)$ and $f'(x)$ don't overlap?

Great observation! If $f(x)$ and $f'(x)$ shared a zero at some point $k$, then $f(k) = 0$ and $f'(k) = 0$. This would imply a point of tangency on the x-axis. However, given the values $f(a)=0, f(b)=2, f(c)=-1, f(d)=2, f(e)=0$, the function must cross the axis to reach these values, and the extrema occur strictly between these crossings. Thus, the sets of zeros are distinct.

Q: Why does $f(x)$ have to have zeros at $x_1$ and $x_2$?

That is the beauty of the Intermediate Value Theorem (IVT). Since $f(b)=2$ (positive) and $f(c)=-1$ (negative), the function must pass through zero to get from positive to negative. Since $f$ is continuous, it cannot 'jump' over the x-axis. Therefore, there must be at least one point $x_1$ in $(b, c)$ where $f(x_1)=0$.

Q: Is it possible for $g(x)$ to have more than 6 zeros?

Absolutely! The question asks for the *minimum* number of zeros. Rolle's Theorem provides a lower bound based on the guaranteed behavior of the function. While the function could oscillate wildly and create many more zeros, we are mathematically certain of at least 6 based on the constraints provided.

Q: I keep getting confused by the notation $f''(x)f(x)$. Is this a second derivative test?

It looks like it, but don't be fooled! The second derivative test is about concavity. Here, the expression is purely algebraic. By recognizing it as the derivative of $f(x)f'(x)$, we bypass the need to analyze concavity entirely and focus on the roots of the function.

Question

For a twice differentiable function $f(x), g(x)$ is defined as $g(x) = (f'(x))^2 + f''(x)f(x)$ on $[a, e]$. If for $a < b < c < d < e, f(a) = 0, f(b) = 2, f(c) = -1, f(d) = 2, f(e) = 0$ then find the minimum number of zeros of $g(x)$.

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Calculus

Welcome, fellow traveler on the path to JEE mastery. Today, we confront a problem that at first glance looks like a chaotic mess of derivatives.

You see $g(x) = (f'(x))^2 + f''(x)f(x)$ and your instinct might be to panic. You might wonder, 'How can I possibly know the behavior of the second derivative $f''(x)$ when I barely know the function $f(x)$?'

But here is the secret: in the world of competitive mathematics, complexity is often a mask for elegance.

### The Detective Work: Unmasking $g(x)$

Let us look at the expression $g(x) = (f'(x))^2 + f''(x)f(x)$ again. Does it not remind you of the Product Rule?

Recall that for any two functions $u(x)$ and $v(x)$, the derivative of their product is:

$$\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$$

If we set $u(x) = f(x)$ and $v(x) = f'(x)$, then $v'(x) = f''(x)$. Substituting these into the product rule gives us exactly:

$$\frac{d}{dx}[f(x)f'(x)] = f(x)f''(x) + (f'(x))^2$$

Suddenly, the 'terrifying' expression $g(x)$ is revealed to be nothing more than the derivative of a simpler auxiliary function, $h(x) = f(x)f'(x)$.

We have transformed our problem: instead of hunting for zeros of $g(x)$, we are hunting for the zeros of the derivative of $h(x)$. By Rolle's Theorem, if we find $n$ zeros of $h(x)$, we are guaranteed at least $n-1$ zeros for $h'(x)$, which is our $g(x)$.

### Mapping the Terrain: The Zeros of $h(x)$

Now, we must find the zeros of $h(x) = f(x)f'(x)$. A product is zero if either factor is zero. So, $h(x) = 0$ whenever $f(x) = 0$ or $f'(x) = 0$.

Let us plot our known points: $f(a)=0, f(b)=2, f(c)=-1, f(d)=2, f(e)=0$.

Using the Intermediate Value Theorem (IVT), we see that because $f(b)=2$ and $f(c)=-1$, there must be a zero $x_1$ in $(b, c)$. Similarly, because $f(c)=-1$ and $f(d)=2$, there must be a zero $x_2$ in $(c, d)$.

Including the endpoints $a$ and $e$, we have at least four zeros for $f(x)$: $a, x_1, x_2, e$.

But we are not done! We also need the zeros of $f'(x)$. Between any two points where $f(x)$ has the same value, the derivative must vanish.

Between $a$ and $x_1$, the function rises from $0$ to $2$ and falls back to $0$, creating a local maximum at some point $c_1$. Between $x_1$ and $x_2$, it dips to $-1$, creating a local minimum at $c_2$. Between $x_2$ and $e$, it rises to $2$ and falls to $0$, creating a local maximum at $c_3$.

These are three distinct points where $f'(x) = 0$.

### The Final Synthesis

We have identified 4 zeros from $f(x)$ and 3 zeros from $f'(x)$. Since the function values at the extrema are non-zero, these 7 points are distinct.

Thus, $h(x)$ has at least 7 zeros. By Rolle's Theorem, the derivative $h'(x)$, which is our $g(x)$, must have at least $7 - 1 = 6$ zeros.

Do you see the beauty here? We didn't need to know the exact shape of $f(x)$. We only needed to understand the fundamental relationship between a function, its derivative, and the intervals defined by its values.

You have successfully navigated the logic of the problem. Keep this mindset—always look for the underlying structure before diving into the algebra. You are ready for the next challenge.

Enter Numerical Value:

Visualized Solution

Analyze the expression for $g (x)$

Identify the Derivative Pattern

Define Auxiliary Function $h (x)$

Plotting the Given Data Points

Zeros of $f (x)$ via IVT

Zeros of $f^{'} (x)$ via Extrema

Total Zeros of $h (x)$

Applying Rolle's Theorem to $h (x)$

Conclusion and Key Takeaway

The Sigma Insight: Mean Value Theorems

The Hidden Symmetry of Calculus

The Detective Work

Mapping the Terrain

The Final Synthesis

Similar Questions

Let $f : R \to R$ be a thrice differentiable function such that $f (0) = 0, f (1) = 1, f (2) = - 1, f (3) = 2$ and $f (4) = - 2$ . Then, the minimum number of zeros of $(3 f^{'} f^{''} + f f^{'''}) (x)$ is _______

Let f and g be twice differentiable even functions on (-2, 2) such that $f (\frac{1}{4}) = 0, f (\frac{1}{2}) = 0, f (1) = 1$ and $g (\frac{3}{4}) = 0, g (1) = 2$ Then, the minimum number of solutions of $f (x) g^{''} (x) + f^{'} (x) g^{'} (x) = 0$ in (-2, 2) is equal to ______.

Let $f : (a, b) \to R$ be twice differentiable function such that $f (x) = \int_{a}^{x} g (t) d t$ for a differentiable function $g (x)$ . If $f (x) = 0$ has exactly five distinct roots in $(a, b)$ , then $g (x) g^{'} (x) = 0$ has at least :

Let $g : R \to R$ be a non constant twice differentiable such that $g^{'} (\frac{1}{2}) = g^{'} (\frac{3}{2})$ . If a real valued function $f$ is defined as $f (x) = \frac{1}{2} [g (x) + g (2 - x)]$ , then

Let $f : [0, 1] \to R$ be a twice differentiable function in $(0, 1)$ such that $f (0) = 3$ and $f (1) = 5$ . If the line $y = 2 x + 3$ intersects the graph of $f$ at only two distinct points in $(0, 1)$ , then the least number of points $x \in (0, 1)$ , at which $f^{''} (x) = 0$ , is ________

If $f (x)$ and $g (x)$ are differentiable function for $0 \leq x \leq 1$ such that $f (0) = 2, g (0) = 0, f (1) = 6, g (1) = 2$ , then show that there exist $c$ satisfying $0 < c < 1$ and $f^{'} (c) = 2 g^{'} (c)$ .

Let $f$ be a twice differentiable function on $(1, 6)$ . If $f (2) = 8$ , $f^{'} (2) = 5$ , $f^{'} (x) \geq 1$ and $f^{''} (x) \geq 4$ for all $x \in (1, 6)$ , then :

Let $f$ be any continuous function on $[0, 2]$ and twice differentiable on $(0, 2)$ . If $f (0) = 0, f (1) = 1$ and $f (2) = 2$ , then

For all twice differentiable functions $f : R \to R$ , with $f (0) = f (1) = f^{'} (0) = 0$

Enter Numerical Value:

Visualized Solution

Analyze the expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Identify the Derivative Pattern

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

Plotting the Given Data Points

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

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

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

Applying Rolle's Theorem to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)