Question

Let $f(x)=\frac{x^3}{3}+\frac{x^2}{2}+x+2$ and $g'(x)=(x^2-9)(x^2-4x+3)(x^2-3x+2)(x^2-2x-3)$.

If $n_1$, $n_2$ and $n_3$ denote number of points of local minima, number of points of local maxima and number of points of inflection of the function $f(g(x))$ then find the value of $(n_1+n_2+n_3)$.

Answer 1

9

Answer 2

To find the number of local minima ($n_1$), local maxima ($n_2$), and inflection points ($n_3$) for the function $h(x) = f(g(x))$, we first analyze $f(x)$ and $g'(x)$.

1. Analyze $f(x)$: $f(x) = \frac{x^3}{3} + \frac{x^2}{2} + x + 2$ $f'(x) = x^2 + x + 1$

The discriminant of $f'(x)$ is $\Delta = 1^2 - 4(1)(1) = -3 < 0$. Since the leading coefficient is positive ($1 > 0$), $f'(x) > 0$ for all $x \in \mathbb{R}$. This means $f(x)$ is a strictly increasing function. Consequently, the local extrema of $f(g(x))$ occur at the same points as the local extrema of $g(x)$, and their nature (maxima/minima) is preserved. Also, $f''(x) = 2x+1$.

2. Analyze $g'(x)$ to find local extrema of $g(x)$: $g'(x) = (x^2-9)(x^2-4x+3)(x^2-3x+2)(x^2-2x-3)$ Factorize each term: $x^2-9 = (x-3)(x+3)$ $x^2-4x+3 = (x-1)(x-3)$ $x^2-3x+2 = (x-1)(x-2)$ $x^2-2x-3 = (x-3)(x+1)$ So, $g'(x) = (x-3)(x+3)(x-1)(x-3)(x-1)(x-2)(x-3)(x+1)$ $g'(x) = (x-3)^3 (x-1)^2 (x-2) (x+1) (x+3)$

The roots of $g'(x)$ are $x = -3, -1, 1, 2, 3$. Let's analyze the sign changes of $g'(x)$ around these roots:

• For $x > 3$: $g'(x) > 0$ (all factors positive).
• As $x$ decreases through $x=3$: $(x-3)^3$ changes sign (odd power). So $g'(x)$ changes from $+$ to $-$. Thus, $x=3$ is a local maximum.
• As $x$ decreases through $x=2$: $(x-2)$ changes sign (odd power). So $g'(x)$ changes from $-$ to $+$. Thus, $x=2$ is a local minimum.
• As $x$ decreases through $x=1$: $(x-1)^2$ does not change sign (even power). So $g'(x)$ changes from $+$ to $+$. Thus, $x=1$ is neither a local maximum nor minimum.
• As $x$ decreases through $x=-1$: $(x+1)$ changes sign (odd power). So $g'(x)$ changes from $+$ to $-$. Thus, $x=-1$ is a local maximum.
• As $x$ decreases through $x=-3$: $(x+3)$ changes sign (odd power). So $g'(x)$ changes from $-$ to $+$. Thus, $x=-3$ is a local minimum.

Summary of local extrema for $g(x)$:

• Local maxima: $x=3, x=-1$. So $n_2 = 2$.
• Local minima: $x=2, x=-3$. So $n_1 = 2$.

3. Find points of inflection for $f(g(x))$: A point $x_0$ is an inflection point of $h(x)=f(g(x))$ if $h''(x_0)=0$ and $h''(x)$ changes sign at $x_0$. $h'(x) = f'(g(x)) g'(x)$ $h''(x) = f''(g(x)) (g'(x))^2 + f'(g(x)) g''(x)$ Since $f'(x) = x^2+x+1$, $f'(g(x)) = g(x)^2+g(x)+1 > 0$ for all $x$. This term is always positive and never zero. Also, $f''(x) = 2x+1$.

The inflection points of $h(x)$ can arise from two types of points: Type A: Points where $g''(x)=0$ and $g''(x)$ changes sign, AND $g'(x) \neq 0$. Type B: Points where $g'(x)=0$ (and $g'(x)$ does not change sign), leading to $h''(x)$ changing sign. Type C: Points where $f''(g(x))=0$ AND $g'(x)=0$ (and $g''(x)$ does not change sign).

Let's find the roots of $g''(x)$. The degree of $g'(x)$ is 8, so the degree of $g''(x)$ is 7. Roots of $g'(x)$ are $-3, -1, 1, 2, 3$.

• If $x_0$ is a root of $g'(x)$ with multiplicity $m$, then $x_0$ is a root of $g''(x)$ with multiplicity $m-1$.
  • For $x=3$, multiplicity is 3. So $x=3$ is a root of $g''(x)$ with multiplicity 2 (even).
  • For $x=1$, multiplicity is 2. So $x=1$ is a root of $g''(x)$ with multiplicity 1 (odd).
  • For $x=-3, -1, 2$, multiplicity is 1. So these are not roots of $g''(x)$ unless $g''(x)$ has other roots at these points, which is not the case from this rule.

By Rolle's Theorem, there is at least one root of $g''(x)$ between any two distinct roots of $g'(x)$. The distinct roots of $g'(x)$ are $-3, -1, 1, 2, 3$. So, there are roots of $g''(x)$ in the intervals:

1. $(-3, -1)$: Let it be $c_1$.
2. $(-1, 1)$: Let it be $c_2$.
3. $(1, 2)$: Let it be $c_3$.
4. $(2, 3)$: Let it be $c_4$. These 4 roots $c_1, c_2, c_3, c_4$ are distinct and different from $-3, -1, 1, 2, 3$. We have found $1$ root at $x=1$ (multiplicity 1) and $2$ roots at $x=3$ (multiplicity 2). Total roots found for $g''(x)$: $c_1, c_2, c_3, c_4, 1, 3, 3$. This accounts for all $4+1+2=7$ roots of $g''(x)$. All $c_i$ roots must be simple roots (multiplicity 1) because the sum of multiplicities equals the degree of $g''(x)$.

Now, let's identify the inflection points for $h(x)$: An inflection point occurs where $h''(x)=0$ and $h''(x)$ changes sign. $h''(x) = (2g(x)+1)(g'(x))^2 + (g(x)^2+g(x)+1)g''(x)$.

Consider the roots of $g''(x)$: $c_1, c_2, c_3, c_4, 1, 3$.

• At $x=1$: $g''(1)=0$. $g'(1)=0$. $h''(1) = (2g(1)+1)(g'(1))^2 + (g(1)^2+g(1)+1)g''(1) = (2g(1)+1)(0)^2 + (g(1)^2+g(1)+1)(0) = 0$. Since $x=1$ is a root of $g''(x)$ with odd multiplicity (1), $g''(x)$ changes sign at $x=1$. Also, since $(x-1)^2$ is a factor of $g'(x)$, $g'(x)$ does not change sign at $x=1$. The term $f'(g(x))g''(x)$ changes sign at $x=1$ (because $f'(g(x)) > 0$ and $g''(x)$ changes sign). The term $f''(g(x))(g'(x))^2$ is $0$ at $x=1$ and is non-negative around $x=1$ (due to $(g'(x))^2$). So, the sign change of $h''(x)$ is dominated by the $f'(g(x))g''(x)$ term. Thus, $x=1$ is an inflection point.
• At $x=3$: $g''(3)=0$. $g'(3)=0$. $h''(3) = (2g(3)+1)(g'(3))^2 + (g(3)^2+g(3)+1)g''(3) = 0$. Since $x=3$ is a root of $g''(x)$ with even multiplicity (2), $g''(x)$ does not change sign at $x=3$. The term $f'(g(x))g''(x)$ does not change sign at $x=3$. The term $f''(g(x))(g'(x))^2$ is $0$ at $x=3$ and is non-negative around $x=3$. Thus, $h''(x)$ does not change sign at $x=3$. So $x=3$ is NOT an inflection point.
• At $x=c_1, c_2, c_3, c_4$: $g''(c_i)=0$ and $g'(c_i) \neq 0$. $h''(c_i) = (2g(c_i)+1)(g'(c_i))^2 + (g(c_i)^2+g(c_i)+1)g''(c_i) = (2g(c_i)+1)(g'(c_i))^2$. For $h''(c_i)$ to be zero, we must have $2g(c_i)+1=0$, which means $g(c_i) = -1/2$. If $g(c_i) \neq -1/2$, then $h''(c_i) \neq 0$, so $c_i$ is not an inflection point. If $g(c_i) = -1/2$, then $h''(c_i)=0$. To be an inflection point, $h''(x)$ must change sign. The term $f''(g(x))(g'(x))^2$ changes sign only if $f''(g(x))$ changes sign, which means $g(x)$ crosses $-1/2$. The term $f'(g(x))g''(x)$ changes sign because $g''(x)$ changes sign at $c_i$ (since $c_i$ are simple roots). In general, for $h''(x) = A(x)B(x) + C(x)D(x)$, if $D(x_0)=0$ and $D(x)$ changes sign, and $B(x_0) \neq 0$, then $h''(x)$ will change sign unless $A(x_0)B(x_0) = 0$ and $A(x)$ or $B(x)$ also changes sign. Since $g'(c_i) \neq 0$, the term $(g'(x))^2$ is positive near $c_i$. The sign of $h''(x)$ near $c_i$ is governed by $f'(g(x))g''(x)$ if $f''(g(x))(g'(x))^2$ does not change sign or is zero. Since $g''(x)$ changes sign at $c_i$ (as they are simple roots), and $f'(g(x)) > 0$, the term $f'(g(x))g''(x)$ changes sign at $c_i$. The term $f''(g(x))(g'(x))^2$ generally does not change sign at $c_i$ unless $g(x)$ crosses $-1/2$ at $c_i$. Even if $g(c_i) = -1/2$, which makes $f''(g(c_i))=0$, the term $f''(g(x))(g'(x))^2$ would change sign if $g(x)$ crosses $-1/2$. If $g(x)$ crosses $-1/2$ at $c_i$, then both terms $f''(g(x))(g'(x))^2$ and $f'(g(x))g''(x)$ change sign. Let $P(x) = f''(g(x))(g'(x))^2$ and $Q(x) = f'(g(x))g''(x)$. $P(x)$ changes sign if $g(x)$ crosses $-1/2$. $Q(x)$ changes sign because $g''(x)$ changes sign. Consider the case where $g(c_i) = -1/2$. Then $f''(g(c_i))=0$. The function $g(x)$ is monotonic between the local extrema. $g'(x)$ sign: $(-\infty, -3): +$; $(-3, -1): -$; $(-1, 1): +$; $(1, 2): +$; $(2, 3): -$; $(3, \infty): +$. So $g(x)$ increases on $(-\infty, -3)$, decreases on $(-3, -1)$, increases on $(-1, 2)$, decreases on $(2, 3)$, increases on $(3, \infty)$. The roots $c_1, c_2, c_3, c_4$ are located in the intervals $(-3, -1)$, $(-1, 1)$, $(1, 2)$, $(2, 3)$. For $c_1 \in (-3, -1)$, $g(x)$ is decreasing. So $g(x)$ is monotonic in this interval. For $c_2 \in (-1, 1)$, $g(x)$ is increasing. For $c_3 \in (1, 2)$, $g(x)$ is increasing. For $c_4 \in (2, 3)$, $g(x)$ is decreasing. Since $g(x)$ is monotonic on each interval containing $c_i$, $g(x)$ can cross $-1/2$ at most once in each interval. It is highly unlikely that $g(x)$ values will be such that the sign changes cancel out precisely at all $c_i$. Assuming general positions of $g(x)$ values, these 4 points ($c_1, c_2, c_3, c_4$) are inflection points. Along with $x=1$, this gives $n_3 = 5$.

Therefore: $n_1 = 2$ (local minima at $x=2, -3$) $n_2 = 2$ (local maxima at $x=3, -1$) $n_3 = 5$ (inflection points at $x=1, c_1, c_2, c_3, c_4$)

The value of $(n_1+n_2+n_3) = 2+2+5 = 9$.