Applications of Derivatives

The absolute maximum value of function $ f(x) = x^3 - 3x + 2 $ in [0, 2] is:

EXPLANATION \[f'(x) = \frac{d}{dx}(x^3 - 3x + 2) = 3x^2 - 3\] Set the derivative equal to zero to find the critical points:\[3x^2 - 3 = 0\]\[3(x^2 - 1) = 0\]\[x^2 = 1\]\[x = \pm 1\]The critical points are $x = -1$ and $x = 1$. Since the interval is $[0, 2]$, we only consider the critical point $x = 1$, as $x = -1$ is outside the interval.2. Evaluate Function at Critical Points and Endpoints

📍The absolute maximum value of a continuous function on a closed interval must occur at a critical point within the interval or at the endpoints of the interval.

We evaluate $f(x)$ at $x=1$, $x=0$, and $x=2$.At the critical point $x=1$:\[f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = \mathbf{0}\]At the left endpoint $x=0$:\[f(0) = (0)^3 - 3(0) + 2 = 0 - 0 + 2 = \mathbf{2}\]At the right endpoint $x=2$:\[f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = \mathbf{4}\]

The largest value is 4.Therefore, the absolute maximum value of the function $f(x) = x^3 - 3x + 2$ in the interval $[0, 2]$ is $\mathbf{4}$.

The function $f(x)=x^{2}-4x+6$ is increasing in the interval

EXPLANATION \[f(x) = x^2 - 4x + 6\] \[f'(x) = 2x - 4\] For Critical Point set $f'(x) = 0$ implies \[2x - 4 = 0 \implies 2x = 4 \implies x = 2\] The function is increasing when $f'(x) > 0$ implies \[2x - 4 > 0\]\[2x > 4\]\[\mathbf{x > 2}\]

A cylindrical tank of radius $10$ cm is being filled with sugar at the rate of $100~\pi~cm^{3}/s$. The rate, at which the height of the sugar inside the tank is increasing, is:

EXPLANATION

The rate at which the height of sugar inside the cylindrical tank increases can be determined using the formula for the volume of a cylinder:

$V = \pi r^2 h$

Given:

Radius, $r = 10 \text{ cm}$
Rate of change of volume, $\dfrac{dV}{dt} = 100\pi \text{ cm}^3/\text{s}$

Since the radius remains constant, differentiate both sides of the volume equation with respect to time $t$:

$\dfrac{dV}{dt} = \pi r^2 \dfrac{dh}{dt}$

Substitute the known values:

$100\pi = \pi (10)^2 \dfrac{dh}{dt}$

$100\pi = 100\pi \dfrac{dh}{dt}$

Dividing both sides by $100\pi$:

$\dfrac{dh}{dt} = 1 \text{ cm/s}$

Therefore, the height of the sugar in the tank is increasing at a rate of $1 \text{ cm/s}$.

The values of $\lambda$ so that $f(x)=\sin x-\cos x-\lambda x+C$ decreases for all real values of x are:

EXPLANATION To determine the values of $\lambda $ for which the function $f(x)=\sin x-\cos x-\lambda x+C$ is always decreasing, we need to find the values of $\lambda $ for which the derivative of $f(x)$ is less than or equal to zero for all real $x$.

The derivative of $f(x)$ with respect to $x$ is:

$f^{\prime }(x)=\cos x+\sin x-\lambda $

For $f(x)$ to be a decreasing function for all real values of $x$, we must have:$f^{\prime }(x)\le 0$$\cos x+\sin x-\lambda \le 0$ that is

$\cos x+\sin x\le \lambda $

Note that for an expression of the form $a\cos x+b\sin x$, the maximum value is $\sqrt{a^{2}+b^{2}}$.

Here, $a=1$ and $b=1$. So the maximum value of $\cos x+\sin x$ is:$\sqrt{1^{2}+1^{2}}=\sqrt{2}$

Now the condition $\cos x+\sin x\le \lambda $ for all real $x$.implies that $\lambda $ must be greater than or equal to the maximum possible value of $\cos x+\sin x$.

Therefore, we must have:$\lambda \ge \sqrt{2}$

If $f(x)=2x+\cos x$, then f(x):

EXPLANATION We differentiate $f(x)$ with respect to $x$:$$f'(x) = 2 - \sin x$$ We know that the range of the sine function is $-1 \le \sin x \le 1$. This means that $f'(x)$ is always positive ($f'(x) > 0$) for all real values of $x$.

Since the first derivative $f'(x)$ is strictly positive for all $x$, the function $f(x) = 2x + \cos x$ is an increasing function throughout its domain. This also rules out (A) and (B) because an increasing function has no local maxima or minima.

The slope of the curve $y=-x^{3}+3x^{2}+8x-20$ is maximum at:

EXPLANATION **Correct Option:** A **Reasoning:** * First derivative: $ \frac{dy}{dx} = -3x^2 + 6x + 8 $ * Set second derivative to zero: $ \frac{d^2y}{dx^2} = -6x + 6 = 0 \implies x = 1 $ * Substitute $ x=1 $ into original equation: $ y = -(1)^3 + 3(1)^2 + 8(1) - 20 = -10 $. The point is (1, -10).

Let $f(x)=|x|$, $x\in R$. Then, which of the following statements is **incorrect**?

EXPLANATION **Correct Option:** D **Reasoning:** * $f(x)=|x|$ has $f(x) \ge 0$ and $f(0)=0$, so minimum at $x=0$. * $ \lim_{x \to \pm\infty} |x| = \infty $, so no maximum value. * $ \lim_{x \to 0} |x| = 0 = f(0) $, so continuous at $x=0$. * LHD at $x=0$ is -1, RHD at $x=0$ is 1. Not differentiable at $x=0$.

A spherical ball has a variable diameter $\frac{5}{2}(3x+1).$ The rate of change of its volume w.r.t. x, when $x=1$, is :

EXPLANATION The diameter of the spherical ball is given by $D = \frac{5}{2}(3x+1)$. The radius of the ball is $r = \frac{D}{2} = \frac{1}{2} \cdot \frac{5}{2}(3x+1) = \frac{5}{4}(3x+1)$. The volume of a sphere is $V = \frac{4}{3}\pi r^3$. Substitute the expression for $r$: $V = \frac{4}{3}\pi \left(\frac{5}{4}(3x+1)\right)^3$ $V = \frac{4}{3}\pi \left(\frac{125}{64}(3x+1)^3\right)$ $V = \frac{125\pi}{48}(3x+1)^3$ To find the rate of change of its volume with respect to $x$, we differentiate $V$ with respect to $x$: $\frac{dV}{dx} = \frac{d}{dx}\left(\frac{125\pi}{48}(3x+1)^3\right)$ $\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^{3-1} \cdot \frac{d}{dx}(3x+1)$ $\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^2 \cdot 3$ $\frac{dV}{dx} = \frac{125\pi \cdot 9}{48}(3x+1)^2$ $\frac{dV}{dx} = \frac{125\pi \cdot 3}{16}(3x+1)^2$ $\frac{dV}{dx} = \frac{375\pi}{16}(3x+1)^2$ Now, we evaluate this rate of change when $x=1$: $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16}(3(1)+1)^2$ $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16}(4)^2$ $\left.\frac{dV}{dx}\right|_{x=1} = \frac{375\pi}{16} \cdot 16$ $\left.\frac{dV}{dx}\right|_{x=1} = 375\pi$ The final answer is $\boxed{\text{375\pi}}$.

Let $f(x)$ be a continuous function on [a, b] and differentiable on (a, b). Then, this function $f(x)$ is strictly increasing in (a, b) if

EXPLANATION The solution is as follows: **Step 1:** Recall the relationship between the derivative of a function and its increasing/decreasing behavior. A function is strictly increasing on an interval if its derivative is strictly positive on that interval. **Step 2:** Analyze the given options in light of this relationship. Option A states $f^{\prime}(x) \lt 0$, which implies the function is strictly decreasing. Option B states $f^{\prime}(x) \gt 0$, which implies the function is strictly increasing. Option C states $f^{\prime}(x) = 0$, which implies the function is constant. Option D is about the function's value, not its rate of change. **Step 3:** Conclude based on the definition of a strictly increasing function. For $f(x)$ to be strictly increasing in $(a, b)$, its derivative $f^{\prime}(x)$ must be greater than 0 for all $x$ in $(a, b)$. The final answer is $\boxed{B}$.

The function $f(x)=x^{3}-3x^{2}+12x-18$ is:

EXPLANATION Here's the step-by-step solution: 1. **Find the first derivative of the function:** $f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 12x - 18) = 3x^2 - 6x + 12$. 2. **Analyze the sign of the first derivative:** The derivative is a quadratic function. To determine its sign, we can find its discriminant. The discriminant of $ax^2 + bx + c$ is $b^2 - 4ac$. For $3x^2 - 6x + 12$, the discriminant is $(-6)^2 - 4(3)(12) = 36 - 144 = -108$. 3. **Interpret the discriminant:** Since the discriminant is negative and the leading coefficient (3) is positive, the quadratic $f'(x) = 3x^2 - 6x + 12$ is always positive for all real values of $x$. 4. **Conclude the function's behavior:** Because $f'(x) > 0$ for all $x \in \mathbb{R}$, the function $f(x)$ is strictly increasing on $\mathbb{R}$. The final answer is $\boxed{B}$.

If the sides of a square are decreasing at the rate of $1.5~cm/s$ the rate of decrease of its perimeter is:

EXPLANATION Let the side of the square be $s$. The perimeter of the square is $P = 4s$. We are given that the rate of decrease of the side is $\frac{ds}{dt} = -1.5~cm/s$. We need to find the rate of decrease of its perimeter, which is $\frac{dP}{dt}$. Differentiate the perimeter equation with respect to time $t$: $\frac{dP}{dt} = \frac{d}{dt}(4s)$ $\frac{dP}{dt} = 4 \frac{ds}{dt}$ Substitute the given value of $\frac{ds}{dt}$: $\frac{dP}{dt} = 4 (-1.5~cm/s)$ $\frac{dP}{dt} = -6~cm/s$ The rate of decrease of its perimeter is $6~cm/s$. The final answer is $\boxed{6~cm/s}$.

The function $f(x)=kx-\sin~x$ is strictly increasing for

EXPLANATION The solution proceeds as follows: **Step 1: Find the derivative of the function.** The derivative of $f(x) = kx - \sin~x$ is $f'(x) = k - \cos~x$. **Step 2: Determine the condition for a strictly increasing function.** For a function to be strictly increasing, its derivative must be strictly positive, i.e., $f'(x) > 0$. **Step 3: Apply the condition to the derivative.** We need $k - \cos~x > 0$ for all $x$. This means $k > \cos~x$ for all $x$. **Step 4: Find the maximum value of $\cos~x$.** The maximum value of $\cos~x$ is 1. **Step 5: Determine the condition for $k$.** For $k > \cos~x$ to hold for all $x$, $k$ must be greater than the maximum value of $\cos~x$. Therefore, $k > 1$. **Final Answer:** The function $f(x)=kx-\sin~x$ is strictly increasing for $k\gt1$. The final answer is $\boxed{k\gt1}$.

The function $f(x)=\frac{x}{2}+\frac{2}{x}$ has a local minima at x equal to:

Given a curve $y=7x-x^{3}$ and x increases at the rate of 2 units per second. The rate at which the slope of the curve is changing, when $x=5$ is:

If f(x)=a(x-cos\~x) is strictly decreasing in R, then 'a' belongs to

The interval in which the function f(x)=2x³+9x²+12x-1 is decreasing, is

Determine the values of x for which $f(x)=\frac{x-4}{x+1}$, $x\ne-1$ is an increasing or a decreasing function.

Surface area of a balloon (spherical), when air is blown into it, increases at a rate of $5\text{ mm}^{2}/\text{s}$. When the radius of the balloon is 8 mm, find the rate at which the volume of the balloon is increasing.

For the curve $y=5x-2x^{3}$ if x increases at the rate of $2\text{ units/s}$, then how fast is the slope of the curve changing when $x=2$?

If $f(x)=x+\frac{1}{x}$, $x\ge1$, show that f is an increasing function.

Find the least value of 'a' so that $f(x)=2x^{2}-ax+3$ is an increasing function on $[2, 4]$.

Find the values of 'a' for which $f(x)=\sin x-ax+b$ is increasing on R.

Find the intervals in which function $f(x)=5x^{\frac{3}{2}}-3x^{\frac{5}{2}}$ is (i) increasing (ii) decreasing.

The area of the circle is increasing at a uniform rate of $2~cm^{2}/sec$. How fast is the circumference of the circle increasing when the radius $r=5$ cm?

The volume of a cube is increasing at the rate of $6~cm^{3}/s.$ How fast is the surface area of cube increasing, when the length of an edge is 8 cm?

Find the interval in which the function $f(x)=x^{4}-4x^{3}+10$ is strictly decreasing.

Show that the function $f(x)=4x^{3}-18x^{2}+27x-7$ has neither maxima nor minima.

Show that $f(x)=e^{x}-e^{-x}+x-tan^{-1}x$ is strictly increasing in its domain.

If M and m denote the local maximum and local minimum values of the function $f(x)=x+\frac{1}{x}(x\ne0)$ respectively, find the value of $(M-m)$

If $f(x)=a(\tan x-\cot x)$, where $a>0$, then find whether $f(x)$ is increasing or decreasing function in its domain.

Find the maximum and minimum values of the function given by $f(x)=5+\sin 2x$.

Show that the function $f(x)=\frac{16\sin x}{4+\cos x}-x$, is strictly decreasing in $(\frac{\pi}{2},\pi)$

Amongst all pairs of positive integers with product as 289, find which of the two numbers add up to the least.

Find the value of 'a' for which $f(x)=\sqrt{3}\sin x-\cos x-2ax+6$ is decreasing in R.

The side of an equilateral triangle is increasing at the rate of 3 cm/s. At what rate its area increasing when the side of the triangle is 15 cm?

Find the absolute maximum and absolute minimum values of the function f given by $f(x)=\frac{x}{2}+\frac{2}{x}$ , on the interval [1, 2].

Find the intervals in which the function $f(x)=\frac{log~x}{x}$ is strictly increasing or strictly decreasing.

The relation between the height of the plant (y cm) with respect to exposure to sunlight is governed by the equation $y=4x-\frac{1}{2}x^{2}$, where x is the number of days exposed to sunlight. (i) Find the rate of growth of the plant with respect to sunlight. (ii) In how many days will the plant attain its maximum height? What is the maximum height?

Find the absolute maximum and absolute minimum of function $f(x)=2x^{3}-15x^{2}+36x+1$ on $[1, 5]$.

The perimeter of a rectangular metallic sheet is 300 cm. It is rolled along one of its sides to form a cylinder. Find the dimensions of the rectangular sheet so that volume of cylinder so formed is maximum.

It is given that function $f(x)=x^{4}-62x^{2}+ax+9$ attains local maximum value at $x=1$ Find the value of 'a', hence obtain all other points where the given function f(x) attains local maximum or local minimum values.

(a) The median of an equilateral triangle is increasing at the rate of 2√3 cm/s. Find the rate at which its side is increasing. OR (b) Sum of two numbers is 5. If the sum of the cubes of these numbers is least, then find the sum of the squares of these numbers.

A technical company is designing a rectangular solar panel installation on a roof using 300 metres of boundary material. The design includes a partition running parallel to one of the sides dividing the area (roof) into two sections.

Let the length of the side perpendicular to the partition be $x$ metres and with parallel to the partition be $y$ metres.,

Using first derivative test, calculate the maximum area the company can enclose with the 300 metres of boundary material, considering the parallel partition.

Find the critical points of the area function. Use second derivative test to determine critical points at the maximum area. Also, find the maximum area.

Write the area of the solar panel as a function of $x$

Write the equation for the total boundary material used in the boundary and parallel to the partition in terms of
x and y..

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Applications of Derivatives

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