The definition of monotonically decreasing function is if a function f : (a,b) → R is said to be monotonically decreasing on (a,b) if x₁ < x₂ ⇒ f(x₁) ≥ f(x₂) ∀ x₁, x₂ ∈ (a,b). Hence, mamatical expression is x₁< x₂ ⇒ f(x₁) ≥ f(x₂) ∀ x₁, x₂ ∈ (a,b).

The meaning of a monotonic function is it eir never decreases or never increases. The condition for a function to be monotonically non-increasing is x₁ < x₂ ⇒ f(x₁) ≥ f(x₂) ∀ x₁, x₂ ∈ (a,b).

f(x) = 3x+10.
f?(x) = 3, which shows 3 > 0 for all x ∈ R.
Thus function f(x) is increasing.

Given that, y=3x⁴+2x³-1
Differentiating w.r.t x, we get
\(\frac{dy}{dx}\)=12x³+6x²
The tangent is parallel to x-axis, which implies that slope \(\frac{dy}{dx}\) is 0.
&re4;12x³+6x²=0
6x² (2x+1)=0
?x=0,-\(\frac{1}{2}\)
If x=0
y=3(0)+2(0)-1=-1
If x=-\(\frac{1}{2}\)
y=3\((-\frac{1}{2})^4+2(-\frac{1}{2})^3-1\)
y=\(\frac{3}{16}-\frac{2}{8}-1\)
y=-\(\frac{15}{16}\)
Hence, points at which tangent is parallel to x-axis is (0,-1) and \((-\frac{1}{2},-\frac{15}{16})\).

Monotonically increasing functions are usually referred to as increasing functions whereas monotonically decreasing functions are usually referred to as decreasing functions.

The Marginal cost is rate of change of revenue w.r.t no. of units produced, we get
\(\frac{dP(x)}{dt}\)=0.8x+2
cost(MC)=\(\frac{dP(x)}{dt}|\)_x=5=0.8x+2=0.8(5)+2=4+2=6.

Let, AB be lamp-post whose foot is A, and B is source of light, and given (AB)? = 12(�) ft.
Let MN denote position of man at time t where (MN)’ = 5ft.
Join BN and produce it to meet AM(produced) at P.
Then length of man’s shadow= (MP)’
Assume, (AM)? = x and (MP)? = y. Then,
(PA)? = (AM)? + (MP)? = x + y
And dx/dt = velocity of man = 3
Clearly, triangles APB and MPN are similar.
Thus, (PM)?/(MN)? = (PA)?/(AB)?
Or, y/5 = (x + y)/12(�)
Or, (25/2)y = 5x + 5y
Or, 3y = 2x
Or, y = (2/3)x
Thus, dy/dt = (2/3)(dx/dt)
As, dx/dt = 2,
= 2/3*3 = 2mph

Let, v cubic cm be volume of cube of cube of side x cm.
Then, v = x³
Thus, dv/dx = d/dx(x³) = 3x²
Now, dv = dv/dx * δx = 3x² * δx
By problem, x = 10 and δx = 0.05
Therefore, approximate error = dv = 3 * 10² * 0.05 = 15 cc cm.
Hence, relative error = dv/v = 15/10³ = 0.015

Given that, x=12 cosec? and y=2 sec??
\(\frac{dx}{d?}\)=-12 cosec ? cot??
\(\frac{dy}{d?}\)=2 tan?? sec??
\(\frac{dy}{dx}\)=\(\frac{dy}{d?}.\frac{d?}{dx}=\frac{2 \,tan?? \,sec??}{-12 \,cosec ? \,cot??}\)=-\(\frac{1}{6} \frac{sin??}{cos^2??} � \frac{cos??}{sin^2???}\) = –\(\frac{cot??}{6}\)
\(\frac{dy}{dx}]_{x=\frac{\pi}{4}}\)=\(-\frac{\frac{cot\pi}{4}}{6}=-\frac{1}{6}\)
Hence, slope of normal at ?=π/4 is
–\(\frac{1}{slope \,of \,tangent \,at \,?=\pi/4}=-\frac{-1}{\frac{1}{6}}\)=6

We know that volume V of a cube is given by
V=x³
Differentiating w.r.t x, we get
\(\frac{dV}{dx}\)=3x²
dV=(\(\frac{dV}{dx}\))Δx=3x² Δx
dV=3x² (6x/100)=0.18x³
Therefore, approximate change in volume is 0.18x³.