Question Video: Estimating the Cubic Root of a Number by Using Linear Approximation Mathematics

Video Transcript

By finding the linear approximation of the function 𝑓 of 𝑥 is equal to the cube root of 𝑥 at a suitable value of 𝑥, estimate the value of the cube root of 1001.

The question wants us to estimate the value of the cube root of 1001. And it wants us to do this by finding a linear approximation of the function 𝑓 of 𝑥 is equal to the cube root of 𝑥 at a suitable value of 𝑥. Let’s start by recalling what a linear approximation of the function 𝑓 of 𝑥 at 𝑥 is equal to 𝑎 is. If our function 𝑓 of 𝑥 is differentiable at 𝑥 is equal to 𝑎, then we can approximate the values of 𝑓 near 𝑥 is equal to 𝑎 by using our tangent line. And we call this the linear approximation 𝑙 of 𝑥. Its equation is equal to 𝑓 of 𝑎 plus 𝑓 prime of 𝑎 times 𝑥 minus 𝑎. And of course, this is just the equation of the tangent line to our function 𝑓 of 𝑥 at 𝑥 is equal to 𝑎.

The first thing we need to decide is our value of 𝑎. And to choose this, remember, we want to use our linear approximation to approximate the cube root of 1001. And remember, our linear approximation is more accurate the closer our input is to 𝑥 is equal to 𝑎. So we want 𝑓 of 𝑎 to be close to the cube root of 1001. In our case, 𝑓 of 𝑥 is the cube root of 𝑥. So 𝑓 of 𝑎 is the cube root of 𝑎.

So if we choose 𝑎 is equal to 1000, then the cube root of 𝑎 is the cube root of 1000. We can easily calculate this value. And we know it’s close to the cube root of 1001. So we’ll take 𝑎 to be 1000. We now need to find 𝑓 evaluated at 𝑎 and 𝑓 prime evaluated at 𝑎.

Let’s start with 𝑓 evaluated at 𝑎. That’s 𝑓 evaluated at 1000, which is equal to the cube root of 1000. And we know that 10 cubed is equal to 1000, so 𝑓 of 𝑎 is equal to 10. We now want to find 𝑓 prime of 𝑎. To do this, we need to differentiate our function 𝑓 of 𝑥, which is the cube root of 𝑥. To do this, we’ll rewrite the cube root of 𝑥 by using our laws of exponents. It’s equal to 𝑥 to the power of one-third.

We can now differentiate 𝑓 of 𝑥 by using the power rule for differentiation. We multiply by the exponent of 𝑥, which is one-third, and reduce this exponent by one. We get 𝑓 prime of 𝑥 is equal to one-third times 𝑥 to the power of negative two over three. We want to use this to find an expression for 𝑓 prime of 𝑎. So we substitute 𝑥 is equal to 1000 into our expression for 𝑓 prime of 𝑥. We get one-third times 1000 to the power of negative two over three.

We’ll evaluate this by using our laws of exponents. First, 𝑎 to the power of negative two over three is equal to the cube root of 𝑎 raised to the power of negative two. So we get one-third times the cube root of 1000 to the power of negative two. And the cube root of 1000 is equal to 10. So we can simplify 𝑓 prime of 1000 to be one-third times 10 to the power of negative two. And we can simplify this by using our laws of exponents. 𝑎 to the power of negative 𝑛 is equal to one divided by 𝑎 to the 𝑛th power. So 10 to the power of negative two is one over 10 squared, which is one over 100. So 𝑓 prime of 1000 is equal to one divided by 300.

We’re now ready to find our linear approximation of the cube root of 𝑥 at 𝑥 is equal to 1000. First, we showed that 𝑓 evaluated at 1000 is equal to 10. Next, we showed that 𝑓 prime of 1000 is equal to one divided by 300. Finally, we multiplied one over 300 by 𝑥 minus 𝑎, which is 𝑥 minus 1000.

Remember, we want to estimate the value of the cube root of 1001. The cube root of 1001 is equal to 𝑓 evaluated at 1001. So we can approximate this by substituting 1001 into our linear approximation. Substituting 𝑥 is equal to 1001 into our linear approximation, we get 10 plus one over 300 times 1001 minus 1000. And 1001 minus 1000 is equal to one. So we get 10 plus one over 300, which we can simplify to give us 3001 divided by 300.

Therefore, by finding the linear approximation of 𝑓 of 𝑥 is equal to the cube root of 𝑥 at 𝑥 is equal to 1000, we’ve shown that the cube root of 1001 is approximately equal to 3001 divided by 300.