Question

prove that the function g(X) = cos (3x+4) is continuous at every real number​

Answer 1

Given, $g(x)=cos(3x+4)$

Let $c$ be any arbitrary point in $\mathbb{R}$

We choose $\varepsilon\:(>0)$

Therefore, $|g(x)-g(c)|$

$=|cos(3x+4)-cos(3c+4)|$

$=2|sin\dfrac{3x+4+3c+4}{2}.sin\dfrac{3c+4-3x-4}{2}|$

$=2|sin\dfrac{3x+3c+8}{2}|.|sin\dfrac{3c-3x}{2}|$

$\le 2|sin\dfrac{3}{2}(c-x)|\quad[\because |sin\dfrac{3x+3c+8}{2}|\le 1]$

$\le 2|\dfrac{3}{2}(c-x)|\quad[\because |sinx|\le |x|]$

$\le |x-c|<\varepsilon$ when $|x-c|<\delta\:(=\varepsilon)$

$\therefore g(x)=cos(3x+4)$ is continuous at $c$ and hence on $\mathbb{R}$.

This completes the proof.

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