find the value of `x`
for which function are identical.
`f(x)=xa n dg(x)=1/(1//x)`
step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
the range of the function f(x) = |x -&nbsp;1| is let, y = (1 – 2 tan x)(5 + 4 sin x)we have to find dy/dxas we can observe that y is a product of two functions say u and v where,u = (1 – 2tan x) and v = (5 + 4sin x)∴ y = uvas we know that to find the derivative of product of two function we apply product rule of differentiation.by product rule, we have –…equation 1as, u = (1 – 2tan x)∴ ⇒ ⇒ …..equation 2 {∵ }as, v = 5 + 4sin x⇒ ⇒ …equation 3 {∵ }∴ from equation 1, we can find dy/dx∴ using equation 2 & 3, we get –⇒ ⇒ ∵ sin x = tan x cos x , so we get –⇒ hence, given:- function f(x) = x3 – 12x2 + 36x + 17theorem:- let f be a differentiable real function defined on an open interval (a,b).(i) if f’(x) > 0 for all , then f(x) is increasing on (a, b)(ii) if f’(x) < 0 for all , then f(x) is decreasing on (a, b)algorithm:-(i) obtain the function and put it equal to f(x)(ii) find f’(x)(iii) put f’(x) > 0 and solve this inequation.for the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.here we have,f(x) = x3 – 12x2 + 36x + 17⇒ ⇒ f’(x) = 3x2 – 24x + 36for f(x) lets find critical point, we must have⇒ f’(x) = 0⇒ 3x2 – 24x + 36 = 0⇒ 3(x2 – 8x + 12) = 0⇒ 3(x2 – 6x – 2x + 12) = 0⇒ x2 – 6x – 2x + 12 = 0⇒ (x – 6)(x – 2) = 0⇒ x = 2, 6clearly, f’(x) > 0 if x < 2 and x > 6and f’(x) < 0 if 2 < x < 6thus, f(x) increases on (–∞, 2) ∪ (6, ∞)and f(x) is decreasing on interval x ∈ (2, 6) y = ex + 1 differentiating both sides of this equation with respect to x, we get: thus, the given function is the solution of the corresponding differential equation. formula :- therefore ,put -2x dx = dt doing integration yields-then finally f(x)=-x cos x + sin xto calculate derivative of the above function f(x) we have to apply formula of derivation of products of two functions-f(x) =-x cos(x)+sin(x)f’(x) =-[x(-sin x)+cos x.1]+cos x ;{by formula d/dx (f.g)=f.g’+g.f’}f’(x) =-(-x sin x) - cos x + cos xf’(x) = x.sin(x) (a) g(x) is a horizontal translation of 2 units to the left then a reflection over the x-axis of f(x)
  

(b) h(x) is a horizontal translation of 3 units to the right then a vertical translation os 4 units up of graph of f(x). 1) p(x) = 1/x2 + 1 if x ≤ 3 it is clear that x= 3 is included, so you have to use this function.2) loo k at table d, and compare the given information with the data written inside the table. the sum of a pmf over its entire domain must equal 1.
so cς(x=1 to ∞)σ(y=1 to ∞)[(1/4^x)(1/3^y)] = 1.
 
ς(i=1 to ∞)[1/n^i] = 1/(n-1). so
 
ς(y=1 to ∞)[(1/3^y)] = 1/(3-1) = 1/2 and
ς(x=1 to ∞)[(1/4^x)] = 1/(4-1) =1/3.
 
so, σ(x=1 to ∞)σ(y=1 to ∞)[(1/4^x)(1/3^y)] = (1/2)*(1/3) = 1/6.
 
to make the sum = 1, then c=6. the value of is we need, principal value of cos-1 x is [0,π]also, cos (2nπ + θ) = cos θ for all n є nso,