Question

If f ( x ) = 3x⁴ + x² + 5 - 3 cos x + 2 sin² x , Then Prove that f ( x ) + f ( -x ) = 2f ( x ).





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Answer 1

$\qquad \qquad\huge \sf༆ Answer ༄$

Here's the solution ~

$\qquad \sf  \dashrightarrow \: f(x) = 3 {x}^{4} + x {}^{2} + 5 - 3 \: cos(x) + 2 \: \sin {}^{2} (x)$

now, let's solve for f ( -x) :

$\qquad \sf  \dashrightarrow \: f( - x) = 3 {( - x)}^{4} +( - x {}^{2}) + 5 - 3 \: cos( - x) + 2 \: ( \sin {}^{} ( - x) ) {}^{2}$

$\qquad \sf  \dashrightarrow \: f( - x) = 3 { x}^{4} + x {}^{2} + 5 - 3 \: cos( x) + 2 \: ( - \sin {}^{} ( x) ) {}^{2}$

$\qquad \sf  \dashrightarrow \: f( - x) = 3 { x}^{4} + x {}^{2} + 5 - 3 \: cos( x) + 2 \: \sin {}^{} ( x) {}^{2}$

Now, let's add both equations ~

$\qquad \sf  \dashrightarrow \: f(x) + f( - x) = 3 {x}^{4} + {x}^{2} + 5 - 3 \: cos(x) + 2 \: {}^{} sin {}^{2} (x) + 3 {x}^{4} +{x}^{2} + 5 - 3 \: \cos(x) + 2 \: \sin {}^{2} (x)$

$\qquad \sf  \dashrightarrow \: f(x) + f( - x) = 2(3 {x}^{4} + x² + 5 - 3 \: \cos(x) + 2 \sin {}^{2} (x) )$

$\qquad \sf  \dashrightarrow \: f(x) + f( - x) = 2(f(x) )$

Answer 2

The sum of the function and its reflection is twice the function

Functions and trigonometry

Given the function expressed as:

f(x) = 3x⁴ + x² + 5 - 3 cos x + 2 sin² x

Its reflection f(-x) is expressed as:

f(-x) = 3(-x)⁴ + (-x)² + 5 - 3 cos (-x) + 2 sin² (-x)

f(-x) = 3x⁴ + x² + 5 - 3 cos x + 2 (-sinx)²
f(-x) = 3x⁴ + x² + 5 - 3 cos x + 2 sin²x

Take the sum of f(x) and f(-x) to have:

f (x) + f (-x) =  3x⁴ + x² + 5 - 3 cos x + 2 sin² x + ( 3x⁴ + x² + 5 - 3 cos x + 2 sin²x )

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

f (x) + f (-x) = 2f(x) proved

Hence the sum of the function and its reflection is twice the function

Learn more on the sum of function here; https://brainly.com/question/25109150