Use linear approximation, i.e. the tangent line, to approximate $\sqrt[3]{1.2}$ as follows: Let $f(x)=\sqrt[3]{x}$. The equation of the tangent line to $f(x)$ at $x=1$ can be written in the form $y=m x+b$ where $m$ is: and where $b$ is: Using this, we find our approximation for $\sqrt[3]{1.2}$ is
Solution
Step 1 ：Let's start by finding the derivative of the function \(f(x)=\sqrt[3]{x}\) at \(x=1\). This will give us the slope of the tangent line at that point, which is the value of \(m\) we're looking for.
Step 2 ：Using the power rule for differentiation, we find that \(f'(x) = \frac{1}{3}x^{-\frac{2}{3}}\). Evaluating this at \(x=1\), we find that \(m = \frac{1}{3}\).
Step 3 ：Next, we find the value of \(b\) by substituting \(x=1\) and \(f(x)=1\) into the equation of the line \(y=mx+b\). This gives us \(b = 1 - m = \frac{2}{3}\).
Step 4 ：So, the equation of the tangent line to \(f(x)\) at \(x=1\) is \(y = \frac{1}{3}x + \frac{2}{3}\).
Step 5 ：Finally, we use the equation of the tangent line to approximate the value of \(\sqrt[3]{1.2}\). Substituting \(x=1.2\) into the equation of the line, we find that \(\sqrt[3]{1.2} \approx 1.067\).
Step 6 ：Final Answer: The slope of the tangent line \(m\) is \(\boxed{\frac{1}{3}}\), the y-intercept \(b\) is \(\boxed{\frac{2}{3}}\), and the approximation for \(\sqrt[3]{1.2}\) is approximately \(\boxed{1.067}\).
