If f : R rightarrow R such that f(x) = 5x - 3 | vedclass

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(C) Given $f(x) = 5x - 3\cos x - 4\sin x$.Find the derivative: $f'(x) = 5 + 3\sin x - 4\cos x$.We know that the range of $a\sin x + b\cos x$ is $[-\sq

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one-one and onto

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(C) Given $f(x) = 5x - 3\cos x - 4\sin x$.Find the derivative: $f'(x) = 5 + 3\sin x - 4\cos x$.We know that the range of $a\sin x + b\cos x$ is $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$.Here,$3\sin x - 4\cos x$ has a range of $[-\sqrt{3^2 + (-4)^2}, \sqrt{3^2 + (-4)^2}] = [-5, 5]$.Therefore,$f'(x) = 5 + (3\sin x - 4\cos x) \geq 5 - 5 = 0$.Since $f'(x) \geq 0$ for all $x \in R$ and $f'(x)$ is not identically zero on any interval,$f(x)$ is a strictly increasing function.As $x \rightarrow \infty$,$f(x) \rightarrow \infty$ and as $x \rightarrow -\infty$,$f(x) \rightarrow -\infty$.Since the function is continuous and strictly increasing,its range is $(-\infty, \infty)$,which is equal to the codomain $R$.Thus,the function is both one-one and onto (bijective).

If $f : R \rightarrow R$ such that $f(x) = 5x - 3\cos x - 4\sin x$,then the function $f(x)$ is

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