Maxima and Minima Questions in English

(D) Given function is $f(x) = \frac{x}{8} + \frac{2}{x}$.
First,find the derivative $f'(x) = \frac{1}{8} - \frac{2}{x^2} = \frac{x^2 - 16}{8x^2}$.
To find critical points,set $f'(x) = 0$,which gives $x^2 - 16 = 0$,so $x = 4$ or $x = -4$.
Since the interval is $[1, 6]$,we only consider $x = 4$.
Now,evaluate $f(x)$ at the critical point and the endpoints of the interval $[1, 6]$:
$f(1) = \frac{1}{8} + \frac{2}{1} = \frac{1}{8} + 2 = \frac{17}{8} = 2.125$.
$f(4) = \frac{4}{8} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1$.
$f(6) = \frac{6}{8} + \frac{2}{6} = \frac{3}{4} + \frac{1}{3} = \frac{9+4}{12} = \frac{13}{12} \approx 1.083$.
Comparing the values $f(1) = \frac{17}{8}$,$f(4) = 1$,and $f(6) = \frac{13}{12}$,the maximum value is $\frac{17}{8}$.

(A) Given the function $f(x) = x^3 e^{-3x}$ for $x > 0$.
To find the maximum value,we first find the derivative $f'(x)$ using the product rule:
$f'(x) = \frac{d}{dx}(x^3) \cdot e^{-3x} + x^3 \cdot \frac{d}{dx}(e^{-3x})$
$f'(x) = 3x^2 e^{-3x} + x^3 (-3 e^{-3x})$
$f'(x) = 3x^2 e^{-3x} (1 - x)$
Setting $f'(x) = 0$ for critical points:
$3x^2 e^{-3x} (1 - x) = 0$
Since $x > 0$ and $e^{-3x} \neq 0$,we have $1 - x = 0$,which gives $x = 1$.
To confirm it is a maximum,we check the sign of $f'(x)$ around $x = 1$. For $x < 1$,$f'(x) > 0$ and for $x > 1$,$f'(x) < 0$.
Thus,$f(x)$ has a local maximum at $x = 1$.
The maximum value is $f(1) = (1)^3 e^{-3(1)} = e^{-3}$.

(A) The displacement is given by $x = t^4 - k t^3$.
The velocity $v$ is the derivative of displacement with respect to time $t$:
$v = \frac{dx}{dt} = 4t^3 - 3kt^2$.
To find the condition for minimum velocity,we find the acceleration $a = \frac{dv}{dt}$:
$a = \frac{dv}{dt} = 12t^2 - 6kt$.
For the velocity to be minimum at $t = 2$,the derivative of velocity with respect to time must be zero at $t = 2$:
$\frac{dv}{dt} \big|_{t=2} = 12(2)^2 - 6k(2) = 0$.
$12(4) - 12k = 0$.
$48 - 12k = 0$.
$12k = 48$.
$k = 4$.
Thus,the value of $k$ is $4$.

(B) To find the point where the slope is maximum,we need to maximize $f'(x)$. Let $g(x) = f'(x) = e^x(\sin x + \cos x)$.
For $g(x)$ to have a maximum,we set $g'(x) = f''(x) = 0$.
$f'(x) = e^x(\sin x + \cos x)$
$f''(x) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x$.
Setting $f''(x) = 0$ gives $2e^x \cos x = 0$. Since $e^x \neq 0$,we have $\cos x = 0$.
In the interval $[0, 2\pi]$,$\cos x = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.
We check the second derivative of $g(x)$,which is $g'(x) = f''(x) = 2e^x \cos x$.
$g''(x) = f'''(x) = 2e^x \cos x - 2e^x \sin x = 2e^x(\cos x - \sin x)$.
At $x = \frac{\pi}{2}$,$g''(\frac{\pi}{2}) = 2e^{\pi/2}(0 - 1) = -2e^{\pi/2} < 0$ (Local maximum).
At $x = \frac{3\pi}{2}$,$g''(\frac{3\pi}{2}) = 2e^{3\pi/2}(0 - (-1)) = 2e^{3\pi/2} > 0$ (Local minimum).
Thus,the slope is maximum at $x = \frac{\pi}{2}$.

(B) Given,$f(x) = x^{4} - 4x^{3} + 4x^{2} + c$.
Evaluating at the boundaries of the interval $(1, 2)$:
$f(1) = 1^{4} - 4(1)^{3} + 4(1)^{2} + c = 1 - 4 + 4 + c = 1 + c$.
$f(2) = 2^{4} - 4(2)^{3} + 4(2)^{2} + c = 16 - 32 + 16 + c = c$.
According to the Intermediate Value Theorem,if $f(1) \cdot f(2) < 0$,then there exists at least one root in the interval $(1, 2)$.
$f(1) \cdot f(2) = (1 + c)c$.
For $f(1) \cdot f(2) < 0$,we require $c(c + 1) < 0$,which implies $c \in (-1, 0)$.
Since $f'(x) = 4x^{3} - 12x^{2} + 8x = 4x(x^{2} - 3x + 2) = 4x(x - 1)(x - 2)$,we observe that $f'(x) = 0$ at $x = 0, 1, 2$.
In the interval $(1, 2)$,$f'(x) < 0$,meaning the function is strictly decreasing.
Since the function is strictly decreasing on $(1, 2)$ and changes sign,it must have exactly one zero in $(1, 2)$ for $c \in (-1, 0)$.

(A) Given $f(x)=1-2x+\int_{0}^{x}e^{(x-t)}f(t)dt$.
Multiplying by $e^{-x}$,we get $e^{-x}f(x) = (1-2x)e^{-x} + \int_{0}^{x}e^{-t}f(t)dt$.
Differentiating with respect to $x$ using Leibniz rule:
$e^{-x}f'(x) - e^{-x}f(x) = -2e^{-x} - (1-2x)e^{-x} + e^{-x}f(x)$.
$f'(x) - f(x) = -2 - 1 + 2x + f(x) \Rightarrow f'(x) - 2f(x) = 2x - 3$.
This is a linear differential equation with integrating factor $I.F. = e^{\int -2 dx} = e^{-2x}$.
$f(x)e^{-2x} = \int (2x-3)e^{-2x} dx = (2x-3)\frac{e^{-2x}}{-2} - \int 2 \cdot \frac{e^{-2x}}{-2} dx = -\frac{2x-3}{2}e^{-2x} - \frac{1}{2}e^{-2x} + C$.
$f(x) = -x + \frac{3}{2} - \frac{1}{2} + Ce^{2x} = 1-x + Ce^{2x}$.
Since $f(0) = 1-2(0) + 0 = 1$,we have $1 = 1-0 + C \Rightarrow C=0$.
So,$f(x) = 1-x$.
Now,$g(x) = \int_{0}^{x} (1-t+2)^{15}(t-4)^6(t+12)^{17} dt = \int_{0}^{x} (3-t)^{15}(t-4)^6(t+12)^{17} dt$.
$g'(x) = (3-x)^{15}(x-4)^6(x+12)^{17} = -(x-3)^{15}(x-4)^6(x+12)^{17}$.
Analyzing the sign of $g'(x)$ around critical points $-12, 3, 4$:
For $x < -12$,$g'(x) < 0$.
For $-12 < x < 3$,$g'(x) > 0$.
For $3 < x < 4$,$g'(x) < 0$.
For $x > 4$,$g'(x) < 0$.
Local minimum at $p = -12$ and local maximum at $q = 3$.
Thus,$|p+q| = |-12+3| = |-9| = 9$.

(C) Given $f(t) = \frac{|t+1|}{t^2}$ for $t < 0$.
For $t \in (-1, 0)$,$f(t) = \frac{t+1}{t^2} = \frac{1}{t} + \frac{1}{t^2}$. Then $f'(t) = -\frac{1}{t^2} - \frac{2}{t^3} = -\frac{t+2}{t^3}$. Since $t < 0$,$t^3 < 0$,so $f'(t) > 0$ for $t \in (-1, 0)$.
For $t \in (-\infty, -1)$,$f(t) = \frac{-(t+1)}{t^2} = -\frac{1}{t} - \frac{1}{t^2}$. Then $f'(t) = \frac{1}{t^2} + \frac{2}{t^3} = \frac{t+2}{t^3}$.
For $f(t)$ to be strictly decreasing,$f'(t) < 0$. Since $t^3 < 0$,we need $t+2 > 0$,i.e.,$t > -2$. Thus,$f(t)$ is strictly decreasing on $(-2, -1)$.
Comparing $(-2, -1)$ with $(2\alpha, \alpha)$,we get $\alpha = -1$.
Now,$g(x) = 2\log_e(x-2) - x^2 + 4x + 1$ for $x > 2$.
$g'(x) = \frac{2}{x-2} - 2x + 4 = \frac{2 - 2x(x-2) + 4(x-2)}{x-2} = \frac{2 - 2x^2 + 4x + 4x - 8}{x-2} = \frac{-2x^2 + 8x - 6}{x-2} = \frac{-2(x^2 - 4x + 3)}{x-2} = \frac{-2(x-3)(x-1)}{x-2}$.
For $x > 2$,$g'(x) = 0$ at $x = 3$.
For $x \in (2, 3)$,$g'(x) > 0$ and for $x > 3$,$g'(x) < 0$. Thus,$g(x)$ has a local maximum at $x = 3$.
The local maximum value is $g(3) = 2\log_e(3-2) - 3^2 + 4(3) + 1 = 2\log_e(1) - 9 + 12 + 1 = 0 - 9 + 13 = 4$.

(A) Given $f(x) = x^{2025} - x^{2000}$.
To find the minimum value,we find the derivative $f'(x) = 2025x^{2024} - 2000x^{1999}$.
Setting $f'(x) = 0$,we get $x^{1999}(2025x^{25} - 2000) = 0$.
Since $x \in [0, 1]$,the critical point is $x = (\frac{2000}{2025})^{1/25} = (\frac{80}{81})^{1/25} = \alpha$.
Evaluating $f(\alpha) = (\frac{80}{81})^{2025/25} - (\frac{80}{81})^{2000/25} = (\frac{80}{81})^{81} - (\frac{80}{81})^{80}$.
$f(\alpha) = (\frac{80}{81})^{80} (\frac{80}{81} - 1) = (\frac{80}{81})^{80} (-\frac{1}{81}) = 80^{80} \cdot 81^{-80} \cdot (-81)^{-1} = 80^{80} \cdot (-81)^{-81}$.
Comparing this with $(80)^{80}(n)^{-81}$,we get $n = -81$.

(A) To find the absolute minimum value of the function $f(x) = x^3 - 18x^2 + 96x$ on the interval $[0, 9]$,we first find the derivative $f'(x)$.
$f'(x) = 3x^2 - 36x + 96$.
Setting $f'(x) = 0$ to find critical points:
$3(x^2 - 12x + 32) = 0$
$3(x - 4)(x - 8) = 0$
So,the critical points are $x = 4$ and $x = 8$.
Now,we evaluate the function $f(x)$ at the critical points and the endpoints of the interval $[0, 9]$:
$f(0) = 0^3 - 18(0)^2 + 96(0) = 0$
$f(4) = 4^3 - 18(4)^2 + 96(4) = 64 - 288 + 384 = 160$
$f(8) = 8^3 - 18(8)^2 + 96(8) = 512 - 1152 + 768 = 128$
$f(9) = 9^3 - 18(9)^2 + 96(9) = 729 - 1458 + 864 = 135$
Comparing these values $(0, 160, 128, 135)$,the absolute minimum value is $0$.

(C) We know that the absolute value function $|x+1|$ is always non-negative,meaning $|x+1| \ge 0$ for all $x \in R$.
Multiplying by $-1$ reverses the inequality: $-|x+1| \le 0$.
Adding $3$ to both sides gives: $-|x+1| + 3 \le 0 + 3$,which simplifies to $f(x) \le 3$.
The maximum value is attained when the term $-|x+1|$ is equal to $0$,which happens at $x = -1$.
Therefore,the maximum value of the function is $3$.