let `f: r->r`
be a twice differentiable function such that `f(x+pi)=f(x)` and `f^(x)+f(x)geq0`
for all `x in rdot`
show that `f(x)geq0`
for all `x in rdot`
step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
?solution 
step 1:
we have to draw the graph of form r to r. 
step 2:
the graph of f is the set of all points (x,f(x)) with x in the domain of f.
we take x in [-4,4]
and we will find the separate values of y for separate intervals.
 
 
 

 
......(1)
 

 

 

using x values along x-axis and along y-axis we get the graph of given function form r to r.
step 3:
the graph is as follow 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. let g(x)=cos x^2 , f(x)=x^1/2 and a and b (a<b) be the root of the quadratic equation 18x^2-9(pi)x+(pi)^2=0 then the area bounded by the curve y=(gof)(x)and the lines x=a , x=b and y=0 is. We are given:  

\(\displaystyle{t}={t}{0}+{\left({\frac{{{k}}}{{{2}}}}\right)}{w}\)  

Subtract t0 from both sides:  

\(\displaystyle{t}-{t}{0}={t}{0}+{\left({\frac{{{k}}}{{{2}}}}\right)}{w}-{t}{0}\)  

\(\displaystyle{t}-{t}{0}={\left({\frac{{{k}}}{{{2}}}}\right)}{w}\)  

Multiply both sides by \(\frac{2}{k}\)  

\(\displaystyle{\left({t}-{t}{0}\right)}{\frac{{{2}}}{{{k}}}}={\left({\frac{{{k}}}{{{2}}}}\right)}{w}\cdot{\left({\frac{{{2}}}{{{k}}}}\right)}\)  

\(\displaystyle{\frac{{{2}{\left({t}-{t}{0}\right)}}}{{{k}}}}={w}\)  

\(\displaystyle{w}={2}\frac{{{t}-{t}{0}}}{{k}}\) 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) formula used/theory.there is some value of ‘x’ for which the f(x) comes to 0, that value of ‘x’ is said to be zero of polynomial f(x)we have,f(x) = 2 – xthen,zero of the polynomial f(x)means f(x) = 02 – x = 0-x = -2x = 2∴ zero of polynomial f(x) is 2conclusion.hence, the zero of the polynomial f(x) is 2 1/
a/ the x-intersect between 2 graphs , x^2+4x-5=0>x = -5 or 1
v = pi { integral of [(4x)^2 - (5-x^2)^2] dx , from x=-5 to x=1 } if , passing through (1, 2), satisfies the differential equation , then f(x) = u/v....i.e. u = 5x² + 4x + 3 and v = 2x² - 2x - 14

now, f '(x) = (u'v - uv')/v²

so, f '(x) = [(10x + 4)(2x² - 2x - 14) - (5x² + 4x + 3)(4x - 2)]/(2x² - 2x - 14)²   

a bit of tidying up with the denominator gives:

f '(x) = -2[9x² + 76x + 25]/(2x² - 2x - 14)²

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