Mix Example of Applications of Derivatives Questions in English

(D) Given $f(x) = x^4 (12 \ln x - 7)$.
First derivative: $f'(x) = 4x^3 (12 \ln x - 7) + x^4 (\frac{12}{x}) = 48x^3 \ln x - 28x^3 + 12x^3 = 48x^3 \ln x - 16x^3 = 16x^3 (3 \ln x - 1)$.
Setting $f'(x) = 0$,we get $3 \ln x = 1 \implies \ln x = 1/3 \implies x = e^{1/3}$.
Since $f''(x) = 48x^2 (3 \ln x - 1) + 16x^3 (3/x) = 144x^2 \ln x - 48x^2 + 48x^2 = 144x^2 \ln x$,at $x = e^{1/3}$,$f''(e^{1/3}) = 144(e^{1/3})^2 (1/3) > 0$,so $x = e^{1/3}$ is a point of minima.
For concavity,we check $f''(x) = 144x^2 \ln x$. In $(0, 1)$,$\ln x < 0$,so $f''(x) < 0$,meaning the graph is concave downwards.
At $x = 1$,$f(1) = 1^4 (12 \ln 1 - 7) = -7$. Since $f''(1) = 144(1)^2 \ln 1 = 0$ and the sign of $f''(x)$ changes at $x=1$,$(1, -7)$ is a point of inflection.
Thus,all statements are true.

(C) The given equation is $2e^{|x|} \tan^{-1}|x| = 1$.
This can be rewritten as $2 \tan^{-1}|x| = e^{-|x|}$.
Let $f(x) = 2 \tan^{-1}|x|$ and $g(x) = e^{-|x|}$.
Since both functions are even,we can analyze the behavior for $x \ge 0$.
For $x \ge 0$,$f(x) = 2 \tan^{-1}x$ and $g(x) = e^{-x}$.
At $x = 0$,$f(0) = 2 \tan^{-1}(0) = 0$ and $g(0) = e^0 = 1$. Thus,$f(0) < g(0)$.
As $x \to \infty$,$f(x) \to 2(\frac{\pi}{2}) = \pi \approx 3.14$ and $g(x) \to 0$.
Since $f(x)$ is a strictly increasing function for $x \ge 0$ and $g(x)$ is a strictly decreasing function for $x \ge 0$,there must be exactly one intersection point for $x > 0$.
Due to symmetry about the $y$-axis,there will also be exactly one intersection point for $x < 0$.
Therefore,there are $2$ solutions in total.

(C) The function is $f(x) = \min (\{x\}, \{e^{-x}\})$.
$1$. Discontinuity: The fractional part function $\{x\}$ is discontinuous at all integers $x = 1, 2, \dots, 9$. At $x=10$,the interval is closed,so we consider the limit from the left. Thus,there are $9$ points of discontinuity $(x=1, 2, \dots, 9)$. So,$C = 9$.
$2$. Non-differentiability: The function is non-differentiable at:
- Points of discontinuity: $9$ points.
- Points where $\{x\} = \{e^{-x}\}$: In each interval $(n, n+1)$,$\{x\} = x-n$ and $\{e^{-x}\} = e^{-x} - \lfloor e^{-x} \rfloor$. Since $e^{-x}$ is strictly decreasing,$\{e^{-x}\}$ is also piecewise defined. For $x \in [0, 10]$,$\{e^{-x}\}$ has discontinuities at $x = -\ln(k)$ for $k \in \mathbb{Z}$.
- Sharp corners: Within each interval $(n, n+1)$,the function switches between $\{x\}$ and $\{e^{-x}\}$. There are $10$ such intervals,and in each,there is exactly one intersection point where the graph has a sharp corner. Thus,there are $10$ such points.
- Total non-differentiable points $D = C + 10 = 9 + 10 = 19$.
Therefore,$C + D = 9 + 19 = 28$.

(A) Given $\int_{a}^{b} f(x) dx = \int_{a}^{b} g(x) dx$,we can write $\int_{a}^{b} (f(x) - g(x)) dx = 0$.
Splitting the integral into three parts based on the intersection points $\alpha$ and $\beta$:
$\int_{a}^{\alpha} (f(x) - g(x)) dx + \int_{\alpha}^{\beta} (f(x) - g(x)) dx + \int_{\beta}^{b} (f(x) - g(x)) dx = 0$.
From the graph,in the interval $(a, \alpha)$,$f(x) > g(x)$,so the area $A_1 = \int_{a}^{\alpha} (f(x) - g(x)) dx$.
In the interval $(\alpha, \beta)$,$g(x) > f(x)$,so the area $A_2 = \int_{\alpha}^{\beta} (g(x) - f(x)) dx = -\int_{\alpha}^{\beta} (f(x) - g(x)) dx$.
In the interval $(\beta, b)$,$f(x) > g(x)$,so the area $A_3 = \int_{\beta}^{b} (f(x) - g(x)) dx$.
Substituting these into the integral equation: $A_1 - A_2 + A_3 = 0$,which implies $A_2 = A_1 + A_3$.
Thus,$\int_{\alpha}^{\beta} (g(x) - f(x)) dx = \int_{a}^{\alpha} (f(x) - g(x)) dx + \int_{\beta}^{b} (f(x) - g(x)) dx$.

(A) Given $f(x) = x \cos^{-1}(-\sin |x|)$. Since $\cos^{-1}(-u) = \pi - \cos^{-1}(u)$,we have $f(x) = x(\pi - \cos^{-1}(\sin |x|))$.
Using $\cos^{-1}(\sin |x|) = \cos^{-1}(\cos(\frac{\pi}{2} - |x|)) = \frac{\pi}{2} - |x|$ for $|x| \in [0, \frac{\pi}{2}]$,we get:
$f(x) = x(\pi - (\frac{\pi}{2} - |x|)) = x(\frac{\pi}{2} + |x|)$.
For $x \geq 0$,$f(x) = x(\frac{\pi}{2} + x) = \frac{\pi}{2}x + x^2$. Thus $f^{\prime}(x) = \frac{\pi}{2} + 2x$.
For $x < 0$,$f(x) = x(\frac{\pi}{2} - x) = \frac{\pi}{2}x - x^2$. Thus $f^{\prime}(x) = \frac{\pi}{2} - 2x$.
Checking differentiability at $x=0$: $LHD = \lim_{x \to 0^-} (\frac{\pi}{2} - 2x) = \frac{\pi}{2}$ and $RHD = \lim_{x \to 0^+} (\frac{\pi}{2} + 2x) = \frac{\pi}{2}$. Since $LHD = RHD$,$f$ is differentiable at $x=0$ and $f^{\prime}(0) = \frac{\pi}{2}$.
Now,for $x \in (-\frac{\pi}{2}, 0)$,$f^{\prime\prime}(x) = -2 < 0$,so $f^{\prime}$ is decreasing.
For $x \in (0, \frac{\pi}{2})$,$f^{\prime\prime}(x) = 2 > 0$,so $f^{\prime}$ is increasing.
Thus,option $A$ is correct.

(N/A) Given equation: $(x-a)^{2}+(y-b)^{2}=c^{2}$
Differentiating both sides with respect to $x$:
$\frac{d}{d x}[(x-a)^{2}]+\frac{d}{d x}[(y-b)^{2}]=\frac{d}{d x}(c^{2})$
$2(x-a) + 2(y-b) \cdot \frac{d y}{d x} = 0$
$\frac{d y}{d x} = -\frac{x-a}{y-b}$ --- $(1)$
Differentiating again with respect to $x$:
$\frac{d^{2} y}{d x^{2}} = -\frac{d}{d x} \left[ \frac{x-a}{y-b} \right] = -\frac{(y-b) \cdot 1 - (x-a) \cdot \frac{d y}{d x}}{(y-b)^{2}}$
Substituting $\frac{d y}{d x}$ from $(1)$:
$\frac{d^{2} y}{d x^{2}} = -\frac{(y-b) - (x-a) \cdot \left( -\frac{x-a}{y-b} \right)}{(y-b)^{2}} = -\frac{(y-b)^{2} + (x-a)^{2}}{(y-b)^{3}} = -\frac{c^{2}}{(y-b)^{3}}$
Now,evaluate the expression:
$\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}} = \frac{\left[1 + \frac{(x-a)^{2}}{(y-b)^{2}}\right]^{\frac{3}{2}}}{-\frac{c^{2}}{(y-b)^{3}}} = \frac{\left[\frac{(y-b)^{2} + (x-a)^{2}}{(y-b)^{2}}\right]^{\frac{3}{2}}}{-\frac{c^{2}}{(y-b)^{3}}} = \frac{\left[\frac{c^{2}}{(y-b)^{2}}\right]^{\frac{3}{2}}}{-\frac{c^{2}}{(y-b)^{3}}}$
$= \frac{\frac{c^{3}}{|y-b|^{3}}}{-\frac{c^{2}}{(y-b)^{3}}} = -c$
Since $-c$ is a constant independent of $a$ and $b$,the proof is complete.

(A) Given $f(x) = x^{3} - 6x^{2} + ax + b$. Since $f(2) = 0$,$8 - 24 + 2a + b = 0 \Rightarrow 2a + b = 16$.
Since $f(4) = 0$,$64 - 96 + 4a + b = 0 \Rightarrow 4a + b = 32$.
Solving these,$a = 8, b = 0$. Thus $f(x) = x^{3} - 6x^{2} + 8x$.
$f^{\prime}(x) = 3x^{2} - 12x + 8$. The roots of $f^{\prime}(x) = 0$ are $x = \frac{12 \pm \sqrt{144 - 96}}{6} = 2 \pm \frac{\sqrt{48}}{6} = 2 \pm \frac{2}{\sqrt{3}}$.
For $(S_1)$: $f^{\prime}(2) = -4$ and $f^{\prime}(4) = 8$. Since $f^{\prime}(x)$ is continuous,by the Intermediate Value Theorem,there exists $x_{1} \in (2, 4)$ such that $f^{\prime}(x_{1}) = -1$ (as $-1 \in (-4, 8)$). Also,$f^{\prime}(x)$ has a root $x_{2} = 2 + \frac{2}{\sqrt{3}} \approx 3.15 \in (2, 4)$. Since $f^{\prime}(3) = 27 - 36 + 8 = -1$,we have $x_{1} = 3 < x_{2} \approx 3.15$. Thus $(S_1)$ is true.
For $(S_2)$: $f$ is decreasing on $(2, x_{4})$ and increasing on $(x_{4}, 4)$ where $x_{4} = 2 + \frac{2}{\sqrt{3}}$. $f(x_{4}) = (2 + \frac{2}{\sqrt{3}})^{3} - 6(2 + \frac{2}{\sqrt{3}})^{2} + 8(2 + \frac{2}{\sqrt{3}}) = -\frac{16}{3\sqrt{3}}$.
$2f^{\prime}(x_{3}) = \sqrt{3}f(x_{4}) = \sqrt{3}(-\frac{16}{3\sqrt{3}}) = -\frac{16}{3} \approx -5.33$. So $f^{\prime}(x_{3}) = -\frac{8}{3} \approx -2.67$. Since $f^{\prime}(x)$ takes all values in $[-4, 8]$ on $(2, 4)$,such $x_{3}$ exists. Thus $(S_2)$ is true.

(C) Given $f(x) = 4 \log_{e}(x-1) - 2x^2 + 4x + 5$ for $x > 1$.
First,find the derivative $f'(x)$:
$f'(x) = \frac{4}{x-1} - 4x + 4 = \frac{4}{x-1} - 4(x-1)$.
For $1 < x < 2$,$(x-1) < 1$,so $\frac{4}{x-1} > 4$,implying $f'(x) > 0$. Thus,$f$ is increasing on $(1, 2)$.
For $x > 2$,$(x-1) > 1$,so $\frac{4}{x-1} < 4$,implying $f'(x) < 0$. Thus,$f$ is decreasing on $(2, \infty)$. (Option $A$ is correct).
Since $f$ increases on $(1, 2)$ to a maximum at $x=2$ where $f(2) = 4 \log_{e}(1) - 2(4) + 4(2) + 5 = 5$,and decreases on $(2, \infty)$ towards $-\infty$,the equation $f(x) = -1$ has exactly two solutions. (Option $B$ is correct).
Check $f(e)$ and $f(e+1)$:
$f(e) = 4 \log_{e}(e-1) - 2e^2 + 4e + 5 \approx 4(0.54) - 2(7.39) + 4(2.718) + 5 \approx 2.16 - 14.78 + 10.87 + 5 = 3.25 > 0$.
$f(e+1) = 4 \log_{e}(e) - 2(e+1)^2 + 4(e+1) + 5 = 4 - 2(e^2 + 2e + 1) + 4e + 4 + 5 = 13 - 2e^2 - 4e - 2 + 4e = 11 - 2e^2 \approx 11 - 14.78 = -3.78 < 0$.
Since $f(e) > 0$ and $f(e+1) < 0$,there is a root in $(e, e+1)$. (Option $D$ is correct).
Calculate $f'(e) - f''(2)$:
$f'(e) = \frac{4}{e-1} - 4(e-1) \approx \frac{4}{1.718} - 4(1.718) \approx 2.33 - 6.87 = -4.54$.
$f''(x) = -\frac{4}{(x-1)^2} - 4$.
$f''(2) = -\frac{4}{(2-1)^2} - 4 = -4 - 4 = -8$.
$f'(e) - f''(2) = -4.54 - (-8) = 3.46 > 0$.
Thus,$f'(e) - f''(2) < 0$ is incorrect. (Option $C$ is incorrect).

(A) Given $f(x) = \sin x + (x^3 - 3x^2 + 4x - 2) \cos x$.
Note that $x^3 - 3x^2 + 4x - 2 = (x-1)^3 + (x-1)$.
So,$f(x) = \sin x + ((x-1)^3 + (x-1)) \cos x$.
Calculating the derivative:
$f'(x) = \cos x + [3(x-1)^2 + 1] \cos x - [(x-1)^3 + (x-1)] \sin x$
$f'(x) = \cos x [1 + 3(x-1)^2 + 1] - (x-1)((x-1)^2 + 1) \sin x$
$f'(x) = \cos x [3(x-1)^2 + 2] + (1-x)((x-1)^2 + 1) \sin x$.
For $x \in (0, 1)$,$\cos x > 0$,$\sin x > 0$,$(1-x) > 0$,and the terms in brackets are positive.
Thus,$f'(x) > 0$ for all $x \in (0, 1)$,which means $f$ is strictly increasing (monotone).
Also,$f(0) = \sin(0) + (-2) \cos(0) = -2 < 0$ and $f(1) = \sin(1) + (1-3+4-2) \cos(1) = \sin(1) > 0$.
Since $f$ is continuous and changes sign on $(0, 1)$,by the Intermediate Value Theorem,$f$ has a zero in $(0, 1)$.
Therefore,both statements $I$ and $II$ are true.

(B) Given $f(x) = 2x + \tan^{-1} x$ and $g(x) = \ln(\sqrt{1+x^2} + x)$ for $x \in [0, 3]$.
First,find the derivatives:
$f'(x) = 2 + \frac{1}{1+x^2}$
$g'(x) = \frac{1}{\sqrt{1+x^2} + x} \cdot \left(\frac{x}{\sqrt{1+x^2}} + 1\right) = \frac{1}{\sqrt{1+x^2} + x} \cdot \frac{x + \sqrt{1+x^2}}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+x^2}}$.
For $x \in [0, 3]$:
$f'(x) \in [2 + \frac{1}{1+3^2}, 2 + \frac{1}{1+0^2}] = [2.1, 3]$.
$g'(x) \in [\frac{1}{\sqrt{1+3^2}}, \frac{1}{\sqrt{1+0^2}}] = [\frac{1}{\sqrt{10}}, 1] \approx [0.316, 1]$.
Since $f'(x) > g'(x)$ for all $x \in [0, 3]$,option $A$ is incorrect.
Both $f(x)$ and $g(x)$ are strictly increasing functions on $[0, 3]$.
Thus,$\max f(x) = f(3) = 6 + \tan^{-1} 3$ and $\max g(x) = g(3) = \ln(3 + \sqrt{10})$.
Since $6 + \tan^{-1} 3 > 6$ and $\ln(3 + \sqrt{10}) \approx \ln(6.16) < 2$,it is clear that $f(3) > g(3)$.
Therefore,option $B$ is correct.

(D) To find $m$,we look for the number of roots of $f(x) = 2^x - x^2 = 0$,which is equivalent to $2^x = x^2$.
From the graph,we can see that the curves $y = 2^x$ and $y = x^2$ intersect at three points: one in the negative region (let it be $\alpha$),one at $x = 2$,and one at $x = 4$. Thus,$m = 3$.
To find $n$,we look for the number of roots of $f'(x) = 2^x \ln 2 - 2x = 0$,which is equivalent to $2^x \ln 2 = 2x$.
From the graph of $y = 2^x \ln 2$ and $y = 2x$,we can see that these two curves intersect at two points. Thus,$n = 2$.
Therefore,$m + n = 3 + 2 = 5$.

(C, A, B) $1.$ $f(x) = 4x^3 + 3x^2 + 2x + 1$. Since $f(-1) = -4+3-2+1 = -2$ and $f(-1/2) = 4(-1/8) + 3(1/4) + 2(-1/2) + 1 = -0.5 + 0.75 - 1 + 1 = 0.25$. Since $f(-1) < 0$ and $f(-1/2) > 0$,the root $s$ lies in $(-1, -1/2)$. Checking the options,$s$ lies in $(-3/4, -1/2)$ is a subset of this,so $(C)$ is correct.
$2.$ $t = |s|$. Since $s \in (-3/4, -1/2)$,$t \in (1/2, 3/4)$. The area $A = \int_0^t (4x^3+3x^2+2x+1) dx = [x^4+x^3+x^2+x]_0^t = t^4+t^3+t^2+t$. For $t=1/2$,$A = 1/16 + 1/8 + 1/4 + 1/2 = 15/16 = 0.9375$. For $t=3/4$,$A = 81/256 + 27/64 + 9/16 + 3/4 = (81+108+144+192)/256 = 525/256 \approx 2.05$. The interval $(21/64, 11/16)$ is $(0.328, 0.6875)$,which is incorrect. Re-evaluating: $f(x)$ is increasing. The area is $A(t)$. For $t \in (0.5, 0.75)$,$A(0.5) = 0.9375$ and $A(0.75) \approx 2.05$. Option $(A)$ is $(0.75, 3)$,which contains the range $(0.9375, 2.05)$. Thus $(A)$ is correct.
$3.$ $f'(x) = 12x^2 + 6x + 2$. $f''(x) = 24x + 6$. $f''(x) = 0$ at $x = -1/4$. For $x > -1/4$,$f''(x) > 0$ (increasing). For $x < -1/4$,$f''(x) < 0$ (decreasing). Thus $(B)$ is correct.

(D) Given $f(x) = \frac{x^2-3x-6}{x^2+2x+4}$.
First,find the derivative $f'(x)$ using the quotient rule:
$f'(x) = \frac{(x^2+2x+4)(2x-3) - (x^2-3x-6)(2x+2)}{(x^2+2x+4)^2}$
$f'(x) = \frac{(2x^3 - 3x^2 + 4x^2 - 6x + 8x - 12) - (2x^3 + 2x^2 - 6x^2 - 6x - 12x - 12)}{(x^2+2x+4)^2}$
$f'(x) = \frac{2x^3 + x^2 + 2x - 12 - (2x^3 - 4x^2 - 18x - 12)}{(x^2+2x+4)^2}$
$f'(x) = \frac{5x^2 + 20x}{(x^2+2x+4)^2} = \frac{5x(x+4)}{(x^2+2x+4)^2}$.
Critical points are $x = 0$ and $x = -4$.
For $x \in (-4, 0)$,$f'(x) < 0$,so $f$ is decreasing. Thus,$f$ is decreasing in $(-2, -1)$,so $(A)$ is $TRUE$.
For $x \in (1, 2)$,$f'(x) > 0$,so $f$ is increasing. Thus,$(B)$ is $TRUE$.
To find the range,evaluate $f(0) = -\frac{6}{4} = -\frac{3}{2}$ and $f(-4) = \frac{16+12-6}{16-8+4} = \frac{22}{12} = \frac{11}{6}$.
As $x \rightarrow \pm \infty$,$f(x) \rightarrow 1$.
The range is $[-\frac{3}{2}, \frac{11}{6}]$.
Since the range is not $R$,$f$ is not onto. Thus,$(C)$ and $(D)$ are $FALSE$.

(C) Given $f(x)=\int_0^x e^{t^2}(t-2)(t-3) dt$.
By the Fundamental Theorem of Calculus,$f^{\prime}(x)=e^{x^2}(x-2)(x-3)$.
To find the critical points,set $f^{\prime}(x)=0$,which gives $x=2$ and $x=3$.
Analyzing the sign of $f^{\prime}(x)$:
For $x < 2$,$f^{\prime}(x) > 0$ (increasing).
For $2 < x < 3$,$f^{\prime}(x) < 0$ (decreasing).
For $x > 3$,$f^{\prime}(x) > 0$ (increasing).
Thus,$f$ has a local maximum at $x=2$ and a local minimum at $x=3$. This confirms $(A)$,$(B)$,and $(D)$ are correct.
Now,find $f^{\prime \prime}(x)$:
$f^{\prime \prime}(x) = \frac{d}{dx} [e^{x^2}(x^2-5x+6)] = e^{x^2}(2x)(x^2-5x+6) + e^{x^2}(2x-5) = e^{x^2}(2x^3-10x^2+12x+2x-5) = e^{x^2}(2x^3-10x^2+14x-5)$.
Let $g(x) = 2x^3-10x^2+14x-5$. Since $g(0) = -5$ and $g(1) = 2-10+14-5 = 1$,by the Intermediate Value Theorem,there exists $c \in (0, 1)$ such that $g(c) = 0$,implying $f^{\prime \prime}(c) = 0$. Thus,$(C)$ is also correct.
Therefore,all options $(A), (B), (C), (D)$ are correct.

(B) $(P)$ Let $P(x) = 10x^3 - 45x^2 + 60x + 35$. Then $P'(x) = 30x^2 - 90x + 60 = 30(x-1)(x-2)$.
Since $P'(x) \leq 0$ for $x \in [1, 2]$,$P(x)$ decreases from $P(1) = 60$ to $P(2) = 55$.
For $f(x) = [P(x)/n]$ to be continuous,the range of $P(x)/n$ must not contain any integer. The range is $[55/n, 60/n]$. The length of this interval is $5/n$. For this to contain no integers,we need $5/n < 1$,so $n > 5$. Also,for the function to be continuous,the values must be between two consecutive integers. Checking $n=9$,range is $[55/9, 60/9] = [6.11, 6.66]$,which contains no integer. Thus,$n=9$ is the minimum value.
$(Q)$ For $g(x)$ to be increasing,$g'(x) = (2n^2 - 13n - 15)(3x^2 + 3) \geq 0$. Since $3x^2+3 > 0$,we need $2n^2 - 13n - 15 \geq 0$. Solving $(2n+2)(n-7.5) \geq 0$,we get $n \geq 7.5$. The smallest natural number is $n=8$.
$(R)$ $h'(x) = n(x^2-9)^{n-1}(2x)(x^2+2x+3) + (x^2-9)^n(2x+2) = (x^2-9)^{n-1} [2nx(x^2+2x+3) + 2(x+3)(x-3)]$. For $x=3$ to be a local minimum,the derivative must change sign from negative to positive. This requires $(x-3)^{n-1}$ to change sign,so $n-1$ must be odd,meaning $n$ is even. The smallest even $n > 5$ is $n=6$.
$(S)$ $l(x) = \sum_{k=0}^4 (\sin|x-k| + \cos|x-k+1/2|)$. The term $\sin|x-k|$ is not differentiable at $x=k$. There are $5$ such points $(k=0, 1, 2, 3, 4)$. Thus,there are $5$ points of non-differentiability.

(D) Given $f(x)=\sqrt{x+\sin x}$.
For $f^{\prime}(x)=0$,we differentiate $f(x)$ with respect to $x$:
$f^{\prime}(x) = \frac{1}{2\sqrt{x+\sin x}} \cdot (1+\cos x) = 0$.
This implies $1+\cos x = 0$,so $\cos x = -1$.
Thus,$x = (2n+1)\pi$ for $n \in \mathbb{Z}$.
Now,we find the value of $f(x)$ at these points:
$f((2n+1)\pi) = \sqrt{(2n+1)\pi + \sin((2n+1)\pi)}$.
Since $\sin((2n+1)\pi) = 0$ for all $n \in \mathbb{Z}$,we have $f((2n+1)\pi) = \sqrt{(2n+1)\pi}$.
Let $x = (2n+1)\pi$ and $y = f(x) = \sqrt{x}$.
Squaring both sides,we get $y^2 = x$,which is the equation of a parabola.

(D) $A. f(x) = 3x^4 - 2x^3 - 6x^2 + 6x + 1$
$f'(x) = 12x^3 - 6x^2 - 12x + 6 = 6(2x^3 - x^2 - 2x + 1) = 6(x-1)(2x-1)(x+1)$
For $x > 1$,$f'(x) > 0$,so $f(x)$ is increasing in $[1, \infty)$. Since $[2, \infty) \subset [1, \infty)$,$f(x)$ is increasing in $[2, \infty)$. Thus,$A \rightarrow V$.
$B. f(x) = x + \frac{1}{x}, x < 0$
$f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$
For $x < -1$,$f'(x) > 0$ (increasing). For $-1 < x < 0$,$f'(x) < 0$ (decreasing).
At $x = -1$,$f'(-1) = 0$ and $f''(x) = \frac{2}{x^3} < 0$ for $x < 0$. Thus,$f(x)$ has a maximum at $x = -1$. Thus,$B \rightarrow II$.
$C. f(x) = x^4(7 - x)^3$
$f'(x) = 4x^3(7-x)^3 - 3x^4(7-x)^2 = x^3(7-x)^2 [4(7-x) - 3x] = x^3(7-x)^2(28 - 7x) = 7x^3(7-x)^2(4-x)$
Critical points are $x = 0, 4, 7$. Checking signs of $f'(x)$ around $x = 4$: for $x < 4$,$f'(x) > 0$; for $x > 4$,$f'(x) < 0$. Thus,$f(x)$ has a maximum at $x = 4$. Thus,$C \rightarrow III$.
$D. f(x) = x^4 + (8-x)^4$
$f'(x) = 4x^3 - 4(8-x)^3 = 4(x^3 - (8-x)^3)$
Setting $f'(x) = 0 \Rightarrow x = 8-x \Rightarrow x = 4$.
$f''(x) = 12x^2 + 12(8-x)^2$. Since $f''(4) = 12(16) + 12(16) > 0$,$f(x)$ has a minimum at $x = 4$. Thus,$D \rightarrow I$.
Correct match: $A-V, B-II, C-III, D-I$.