1. Find the second derivative \[\mbox{Find the first derivative using the power rule}\] \[f'\left(x\right)=4x+3\] \[\mbox{Find the second derivative using the power rule}\] \[f''\left(x\right)=4\]
2. Set the derivative equal \(0\) and solve \[f''\left(x\right)=4=0\] \[4\neq 0\]

\[\therefore\]

The function \(f\) has no inflection points.

1. Find the second derivative \[\mbox{Find the first derivative using the power rule}\] \[f'\left(x\right)=3x^{2}+6x+1\] \[\mbox{Find the second derivative using the power rule}\] \[f''\left(x\right)=6x+6\]
2. Set the derivative equal \(0\) and solve \[f''\left(x\right)=6x+6=0\] \[6x+6=0\] \[6x=-6\] \[x=-1\]

\[\therefore\]

The function \(f\) has an inflection point at \(x=-1\)

1. Find the second derivative \[\mbox{Find the first derivative using the power rule}\] \[f'\left(x\right)=\frac{1}{3}x^{3}+2x^{2}+3x+2\] \[\mbox{Find the second derivative using the power rule}\] \[f''\left(x\right)=x^{2}+4x+3\]
2. Set the derivative equal \(0\) and solve \[f''\left(x\right)=x^{2}+4x+3=0\] \[\mbox{Factor the function and solve}\] \[\left(x+1\right)\left(x+3\right)=0\] \[x=-1 \mbox{ or } x=-3\]

\[\therefore\]

The function \(f\) has an inflection point at \(x=-1\) and \(x=-3\)

1. Find the second derivative \[\mbox{Find the first derivative using the chain rule}\] \[f'\left(x\right)=2\left(\cos\left(x\right)\right)\left(\frac{d}{dx}\left[\cos\left(x\right)\right]\right)=\] \[2\cos\left(x\right)\left(-\sin\left(x\right)\right)=-2\sin\left(x\right)\cos\left(x\right)\] \[\mbox{Find the second derivative using the product rule}\] \[f''\left(x\right)=\left[-2\sin\left(x\right)\cdot\frac{d}{dx}\left[\cos\left(x\right)\right]\right]-\left[2\frac{d}{dx}\left[\sin\left(x\right)\right]\cdot\cos\left(x\right)\right]=\] \[\left[-2\sin\left(x\right)\cdot-\sin\left(x\right)\right]-2\left[\cos\left(x\right)\cdot\cos\left(x\right)\right]=\] \[2\sin^{2}\left(x\right)-2\cos^{2}\left(x\right)\]
2. Set the derivative equal \(0\) and solve for values in the given interval \[2\sin^{2}\left(x\right)-2\cos^{2}\left(x\right)=0\] \[\mbox{Rearrange the function}\] \[2\sin^{2}\left(x\right)=2\cos^{2}\left(x\right)\] \[\mbox{Divide both sides by \(2\cos^{2}\left(x\right)\) and simplify}\] \[\frac{2\sin^{2}\left(x\right)}{2\cos^{2}\left(x\right)}=\frac{2\cos^{2}\left(x\right)}{2\cos^{2}\left(x\right)}\] \[\tan^{2}\left(x\right)=1\] \[\tan\left(x\right)=\pm1\] \[x=\tan\left(pm1\right)\] The tangent function is equal to \(1\) at \(\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4},\frac{9\pi}{4}...\); therefore, the tangent function is equal to \(1\) at \(\frac{n\pi}{4}\) where \(n\) is any odd number. Lastly, we have to find the exact values of \(\frac{n\pi}{4}\) that lie in the given interval. Those values are: \[x=\frac{\pi}{4}\mbox{, }x=\frac{3\pi}{4}\mbox{, }x=\frac{5\pi}{4}\mbox{, and }x=\frac{7\pi}{4}\]

\[\therefore\]

The function \(f\) has inflections point at \(x=\frac{\pi}{4}\mbox{, }x=\frac{3\pi}{4}\mbox{, }x=\frac{5\pi}{4}\mbox{, and }x=\frac{7\pi}{4}\)