Question

Another calculus problem

Answer 1

Let f(x) = \[\sqrt[3]{x}\].The equation of the tangent line to f(x) at x = 64 is y =
Using this, we find our approximation for \[\sqrt[3]{64.1}\] is

Answer 2

21.37

Answer 3

The equation of the tangent line at \(x=a\) is given by \(y = f^{\prime}(a)(x-a)+f(a)\). If \(f(x)=\sqrt[3]{x}\), then \(f^{\prime}(x)=\dfrac{1}{3x^{2/3}}\). What are \(f(64)\) and \(f^{\prime}(64)\)? If you can answer this, then you have the key pieces needed to write the equation of your tangent line. Also note that \(64=4^3\); it will make some of your calculations easier to do by hand. :-)

Once you have the tangent line, evaluate it at \(x=64.1\).

Does this make sense?

Answer 4

Kinda @ChristopherToni

Answer 5

That's fine; what part would you like me to explain more? :-)

Answer 6

when i plug in cuberoot(3) for all the A's @ChristopherToni

Answer 7

Since you're finding the tangent line at x=64, we need to know \(f(64)\) and \(f^{\prime}(64)\).

Since \(f(x)=\sqrt[3]{x}\), we have that \(f(64) = \sqrt[3]{64} = \sqrt[3]{4^3} = 4\).

Since \(f^{\prime}(x)=\dfrac{1}{3x^{2/3}}\), we have that \(f^{\prime}(64) = \dfrac{1}{3(64)^{2/3}} = \dfrac{1}{3(4^3)^{2/3}} = \dfrac{1}{3(4^2)} = \dfrac{1}{48}\).

Therefore, the equation of the tangent line is

\(\begin{aligned}y &= f^{\prime}(64)(x-64)+f(64)\\ &= \dfrac{1}{48}(x-64) + 4\\ &= \frac{1}{48}x-\frac{64}{48} + 4 \\ &= \frac{1}{48}x-\frac{4}{3}+4\\ &= \frac{1}{48}x+\frac{8}{3} \end{aligned}\).

To approximate \(\sqrt[3]{64.1}\), plug this into the equation for the tangent line to get

\(y(64.1) = \dfrac{1}{48}(64.1)+\dfrac{8}{3} \approx 4.0021\) (rounded to 4 decimal places).

Does this make sense?

Answer 8

holy pellet thank you for the effort sir

Answer 9

Thank you so much @ChristopherToni

Answer 10

You're welcome! :-)