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Let f[x]=| x-pie/2|^3 +sin^2x then find the value of the fiven function f’’[pie/2]

Let f[x]=| x-pie/2|^3 +sin^2x then find the value of the fiven function f’’[pie/2]

10 months ago

Answers : (3)

Saurabh Koranglekar
askIITians Faculty
10341 Points
							Dear student

Please ask complete question in standard notation or attach the image of the question

Regards
10 months ago
Vikas TU
14149 Points
							
Dear student 
Question is not clear 
Please attach an image, 
We will happy to  help you 
Good Luck
Cheers
10 months ago
Aditya Gupta
2075 Points
							
dear meghana, we see that f(x)= g(x) + h(x) where g(x)= |x – pi/2|^3 and h(x)= sin^2x 
so, f”(x)= g”(x) + h”(x)
now, h’(x)= 2sinxcosx= sin2x so h”(x)= 2cos2x so that h”(pi/2)= 2cospi= – 2
now, to find g”(x) at pi/2, we first find RHL and then LHL: 
RHL= Lt x tends to pi/2+ [g’(x) – g’(pi/2)]/(x – pi/2) and LHL= Lt x tends to pi/2- [g’(x) – g’(pi/2)]/(x – pi/2) 
now, g’(pi/2+)= Lt x tends to pi/2+ [g(x) – g(pi/2)]/(x – pi/2)= Lt x tends to pi/2+ (x – pi/2)^3/(x – pi/2)
= Lt x tends to pi/2+ (x – pi/2)^2= 0
similarly g’(pi/2-)= Lt x tends to pi/2– [g(x) – g(pi/2)]/(x – pi/2)= Lt x tends to pi/2- (pi/2 – x)^3/(x – pi/2)
= Lt x tends to pi/2- – (x – pi/2)^2= 0
hence, g’(pi/2)= 0 since g’(pi/2+)= g’(pi/2-)= 0
so, RHL=  Lt x tends to pi/2+ [g’(x) – 0]/(x – pi/2)=  Lt x tends to pi/2+ g’(x)/(x – pi/2)
but for x greater than pi/2, g’(x)=  Lt h tends to 0 [g(x+h) – g(x)]/h
= Lt h tends to 0 [(x+h – pi/2)^3 – (x – pi/2)^3]/h= 3(x – pi/2)^2
so RHL= Lt x tends to pi/2+ g’(x)/(x – pi/2)= Lt x tends to pi/2+ 3(x – pi/2)^2/(x – pi/2)= 0
similarly LHL= Lt x tends to pi/2- g’(x)/(x – pi/2)
for x less than pi/2, g’(x)= Lt h tends to 0 [g(x+h) – g(x)]/h= Lt h tends to 0 [(pi/2 – x – h)^3 – (pi/2 – x)^3]/h= 3(pi/2 – x)^2*( – 1)= – 3(x – pi/2)^2
so, LHL= Lt x tends to pi/2- g’(x)/(x – pi/2)= LHL= Lt x tends to pi/2- – 3(x – pi/2)^2/(x – pi/2)= 0
since LHL= RHL= 0, we have g”(pi/2)= 0
hence, f”(pi/2)= g”(pi/2) + h”(pi/2)
= 0 + ( – 2)
– 2
KINDLY APPROVE :))
10 months ago
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