Question

Find the range of the function
i) $f\left(x\right)=x^2+2x+3$
ii) $f\left(x\right)=3\sin x+4\cos x-5$

Answer 1

$\left(1\right)$ $f\left(x\right)=y=x^2+2x+3$

$y=(x+1{)}^2+2$

So $y_{min}=2$

Thus range of $y=[2,\infty )$.

$\left(2\right)$ $f\left(x\right)=y=3\sin x+4\cos x-5$

since for $a\sin x+b\cos x$

$-\sqrt{a^2+b^2}\le$$a\sin x+b\cos x\le$$\sqrt{a^2+b^2}$

$-\sqrt{3^2+4^2}\le$$3\sin x+4\cos x\le$$\sqrt{3^2+4^2}$

$-5\le$$3\sin x+4\cos x\le$$5$

$-10\le$$3\sin x+4\cos x-5\le$$0$

Thus range of $y=[-10,0]$.