First, notice that \(\pi\) is approximately equal to \(3\), which means that \(\frac{\pi}{4} \approx \frac{3}{4}\). Therefore, using the tangent line for \(x=\frac{\pi}{4}\), we can get a good approximation for \(\sin\left(\frac{3}{4}\right)\).

‍

1. Find the derivative \[f'\left(x\right)=\cos\left(x\right)\]
2. Find and write the equation of the tangent line \[\mbox{Find \(f\left(\frac{\pi}{4}\right)\) and \(f'\left(\frac{\pi}{4}\right)\)}\] \[f\left(\frac{\pi}{4}\right)=\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\] \[f'\left(\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\] \[\mbox{Write the equation of the tangent line}\] \[y-\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\left(x-\frac{\pi}{4}\right)\] \[y=\frac{\sqrt{2}}{2}\left(x-\frac{\pi}{4}\right)+\frac{\sqrt{2}}{2}=\frac{\sqrt{2}x}{2}-\frac{\sqrt{2}\pi}{8}+\frac{\sqrt{2}}{2}\] \[=\frac{\sqrt{2}x+\sqrt{2}}{2}-\frac{\sqrt{2}\pi}{8}\] \[=\frac{4\sqrt{2}x+4\sqrt{2}-\sqrt{2}\pi}{8}\]
3. Approximate \(\sin\left(\frac{3}{4}\right)\) \[\]We can do this by substituting \(\frac{3}{4}\) into the tangent line equation. \[f\left(\frac{3}{4}\right)\approx\frac{3\sqrt{2}+4\sqrt{2}-\sqrt{2}\pi}{8}=\frac{7\sqrt{2}-\sqrt{2}\pi}{8}\] \[=\frac{\left(7-\pi\right)\sqrt{2}}{8}\]

‍

If you are allowed to do so, we can check our answer with a calculator.

\[\sin\left(\frac{3}{4}\right)=0.6816\]

\[\frac{\left(7-\pi\right)\sqrt{2}}{8}=0.68207\]

As you can see, our approximation is fairly close to the real value.

First, notice that \(e\) is approximately equal to \(3\); therefore, using the tangent line for \(x=3\), we can get a good approximation for \(f\left(3\right)\).

‍

1. Find the derivative \[f'\left(x\right)=\frac{1}{x}\]
2. Find and write the equation of the tangent line \[\mbox{Find \(f\left(e\right)\) and \(f'\left(e\right)\)}\] \[f\left(e\right)=\ln\left(e\right)+1=2\] \[f'\left(e\right)=\frac{1}{e}\] \[\mbox{Write the equation of the tangent line}\] \[y-2=\frac{1}{e}\left(x-e\right)\] \[y=\frac{x}{e}-1+2=\frac{x}{e}+1\]
3. Approximate \(f\left(3\right)\) \[\]We can do this by substituting \(3\) into the tangent line equation. \[f\left(3\right)\approx\frac{3}{e}+1\]

‍

If you are allowed to do so, we can check our answer with a calculator.

\[f\left(3\right)=\ln\left(3\right)+1=2.0986\]

\[\frac{3}{e}+1=2.1036\]

As you can see, our approximation is fairly close to the real value.

1. Find the derivative \[f'\left(x\right)=2x+3\]
2. Find and write the equation of the tangent line \[\mbox{Find \(f\left(0\right)\) and \(f'\left(0\right)\)}\] \[f\left(0\right)=4\] \[f'\left(0\right)=3\] \[\mbox{Write the equation of the tangent line}\] \[y-4=3\left(x-0\right)\] \[y=3x+4\]
3. Approximate \(f\left(0.1\right)\) \[\]We can do this by substituting \(0.1\) into the tangent line equation. \[f\left(0.1\right)\approx3\left(0.1\right)+4=4.3\]

‍

If you are allowed to do so, we can check our answer with a calculator.

\[f\left(0.1\right)=\left(0.1\right)^{2}+3\left(0.1\right)+4=4.31\]

As you can see, our approximation is fairly close to the real value.

This problem is essentially asking us to find the tangent line at \(x=-2\).

1. Find the derivative \[f'\left(x\right)=3x^{2}+6x+2\]
2. Find \(f\left(-2\right)\) and \(f'\left(-2\right)\) \[f\left(-2\right)=\left(-2\right)^{3}+3\left(-2\right)^{2}+2\left(-2\right)+1=\] \[-8+12-4+1=1\] \[f'\left(-2\right)=3\left(-2\right)^{2}+6\left(-2\right)+2=12-12+2=2\]
3. Write the equation of the tangent line \[y-1=2\left(x+2\right)\] \[y=2x+2+1=2x+3\]

The following formula would be used to calculate a local linear approximation of \(f\left(x\right)=\frac{1}{x^{2}+1}\) around \(x=1\)?.

\[y=2x+2+1=2x+3\]

This problem is essentially asking us to find the tangent line at \(x=3\).

1. Find the derivative \[\mbox{Rewrite the expression}\] \[f\left(x\right)=\left(x^{2}+1\right)^{-1}\] \[\mbox{Find the derivative}\] \[f'\left(x\right)=-1\left(2x\right)\left(x^{2}+1\right)^{-2}=\] \[-\frac{2x}{\left(x^{2}+1\right)^{2}}\]
2. Find \(f\left(1\right)\) and \(f'\left(1\right)\) \[f\left(1\right)=\frac{1}{1+1}=\frac{1}{2}\] \[f'\left(1\right)=-\frac{2}{\left(1+1\right)^{2}}=-\frac{1}{2}\]
3. Write the equation of the tangent line \[y-\frac{1}{2}=-\frac{1}{2}\left(x-1\right)\] \[y=-\frac{1}{2}x+\frac{1}{2}+\frac{1}{2}=-\frac{1}{2}x+1\]

The following formula would be used to calculate a local linear approximation of \(f\left(x\right)=\frac{1}{x^{2}+1}\) around \(x=1\)?.

\[y=-\frac{1}{2}x+1\]

First, notice that \(\pi\) is approximately equal to \(3\), which means that \(-\frac{\pi}{2} \approx -\frac{3}{2}\). Therefore, using the tangent line for \(f\left(-\frac{\pi}{2}\right)\), we can get a good approximation for \(f\left(-\frac{3}{2}\right)\).

1. Find the derivative \[f'\left(x\right)=2\left(-\sin\left(x\right)\right)\cos\left(x\right)-\cos\left(x\right)=\] \[-2\sin\left(x\right)\cos\left(x\right)-\cos\left(x\right)\]
2. Find and write the equation of the tangent line \[\mbox{Find \(f\left(-\frac{\pi}{2}\right)\) and \(f'\left(-\frac{\pi}{2}\right)\)}\] \[f\left(\frac{\pi}{2}\right)=\cos^{2}\left(-\frac{\pi}{2}\right)-\sin\left(-\frac{\pi}{2}\right)=0+1=1\] \[f'\left(\frac{\pi}{2}\right)=-2\sin\left(-\frac{\pi}{2}\right)\cos\left(-\frac{\pi}{2}\right)-\cos\left(-\frac{\pi}{2}\right)=0\] \[\mbox{Write the equation of the tangent line}\] \[y-1=0\left(x-\frac{\pi}{2}\right)\] \[y=1\]
3. Approximate \(f\left(-\frac{3}{2}\right)\) \[f\left(-\frac{3}{2}\right)\approx 1\]
4. Use your calculator to find the actual value \[f\left(-\frac{3}{2}\right)=1.0024\]
5. Use the percent error formula to find the error \[\%\mbox{ error}=\left|\frac{1.0024-1}{1.0024}\right|\cdot100\%=0.2394\%\]

As you can see, since our error is so small, our approximation is fairly accurate.

1. Find the derivative \[f'\left(x\right)=\frac{1}{2}\left(\frac{1}{x}\right)\left(\ln\left(x\right)\right)^{-\frac{1}{2}}=\frac{1}{2x\sqrt{\ln\left(x\right)}}\]
2. Find and write the equation of the tangent line \[\mbox{Find \(f\left(e\right)\) and \(f'\left(e\right)\)}\] \[f\left(e\right)=\sqrt{\ln\left(e\right)}=1\] \[f'\left(e\right)=\frac{1}{2e\sqrt{\ln\left(e\right)}}=\frac{1}{2e}\] \[\mbox{Write the equation of the tangent line}\] \[y-1=\frac{1}{2e}\left(x-e\right)\] \[y=\frac{x}{2e}-\frac{1}{2}+1=\frac{x}{2e}+\frac{1}{2}\]
3. Use your calculator to approximate \(f\left(3\right)\) \[f\left(3\right)approx\frac{3}{2e}+\frac{1}{2}=1.0518\]
4. Use your calculator to find the actual value \[f\left(3\right)=\sqrt{\ln\left(3\right)}=1.0481\]
5. Use the percent error formula to find the error \[\%\mbox{ error}=\left|\frac{1.0481-1.0518}{1.0481}\right|\cdot100\%=0.3530\%\]

As you can see, since our error is so small, our approximation is fairly accurate.