Question

F(x)=x³-3x+sinx
show that f(x) + f(-x)=0​

Answer 1

Answer: + 02 - 23 % + 34

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Given :-

f(x)=x^3-3x+sin x

To find:-

Show that f(x)+f(-x) = 0

Solution:-

Given that

f(x) = x^3 -3x + Sin x

put x = -x then

f(-x) = (-x)^3-3(-x)+Sin (-x)

=>f(-x)= -x^3+3x - Sin x

now

f(x)+f(-x)

=>(x^3 -3x + Sin x)+(-x^3+3x - Sin x)

=>x^3-3x+Sin x-x^3+3x - Sin x

=>(x^3-x^3)+(-3x+3x)+(Sin x-Sin x)

=>0+0+0

=>0

f(x)+f(-x) = 0

Hence, Proved