Video: Finding the Linear Approximation of a Cubic Root Function

Find the linear approximation of the function 𝑓(π‘₯) = βˆ›π‘₯ at π‘₯ = βˆ’8.

06:39

Video Transcript

Find the linear approximation of the function 𝑓 of π‘₯ equals the cube root of π‘₯ at π‘₯ equals negative eight.

We have a function, the cube root function, and we know some values that that function takes. The cube root of one is one because one cubed is one. The cube root of eight is two because two cubed is eight. And the cube root of negative one twenty-seventh is negative a third because negative a third cubed is negative one twenty-seventh.

But how do you find the cube root of negative seven? There doesn’t seem to be an easy way without using a calculator. But how does your calculator find the cube root of negative seven. One way is to find a linear approximation of our function, the cube root function. We approximate 𝑓 of π‘₯ by the linear function which takes π‘₯ to π‘šπ‘₯ plus 𝑏 for some numbers π‘š and 𝑏. We can then use this linear function to find an approximate value for the cube root of negative seven.

But how do we find a linear approximation of our function? Let’s take a look at its graph. Here is a graph of our cube root function. We want to find a linear approximation of this function at π‘₯ equals negative eight; that is, a linear function which is very close to our function 𝑓 of π‘₯ equals the cube root of π‘₯ for values of π‘₯ near negative eight. The graph of this linear function will be a straight line.

And as the functions are close near π‘₯ equals negative eight, this straight line will be very close to the graph of our function at π‘₯ equals negative eight. Can we think of what straight line this might be? A good choice is the tangent to our graph at π‘₯ equals negative eight. The exact value of 𝑓 of negative seven is the 𝑦-coordinate of the point on the graph of 𝑓 of π‘₯ with π‘₯-coordinate negative seven.

And we can see from the graph that this is very close to the 𝑦-coordinate of the point on the tangent at π‘₯ equals negative seven. We are, therefore, looking for the equation of the tangent. How do we find the equation of the tangent? Well we can differentiate to find the slope of the tangent. This slope is 𝑓 prime of negative eight. And the tangent passes through the point negative eight 𝑓 of negative eight, which is where the tangent touches the graph.

It’s therefore easy to find the equation of this line in point-slope form. The equation of the line with slope π‘š which passes through the point π‘₯ naught, 𝑦 naught is 𝑦 equals π‘š times π‘₯ minus π‘₯ naught plus 𝑦 naught. And so as the slope π‘š is 𝑓 prime of negative eight, π‘₯ naught is negative eight and 𝑦 naught is 𝑓 of negative eight. We find that 𝑦 is 𝑓 prime of negative eight times π‘₯ minus negative eight plus 𝑓 of negative eight. And remember, we’re using the equation of this tangent to approximate 𝑓 of π‘₯ near π‘₯ equals negative eight.

Okay let’s now clear some room. We just need to find the values of 𝑓 prime of negative eight and 𝑓 of negative eight now. Let’s start with 𝑓 of negative eight. This is the cube root of negative eight. And this is equal to negative two, as negative two cubed is negative eight. 𝑓 prime of negative eight is slightly harder to find. We have to differentiate 𝑓. So we have to differentiate the cube root of π‘₯. We can write the cube root of π‘₯ as π‘₯ to the power of a third, which allows us to apply our rule for the derivative of π‘₯ to the 𝑛 with respect to π‘₯ which works not just when 𝑛 is a whole number or integer but for any real value of 𝑛.

Doing so, we bring the exponent, a third, down and then reduce the exponent by one, so a third minus one is negative two-thirds. 𝑓 prime of negative eight is then a third times negative eight to the power of negative two- thirds, which is a third times one over negative eight to the two-thirds. Using what we know about negative exponents, using more exponent laws, this is a third times one over negative eight to the third squared.

Negative eight to the power of a third is the cube root of negative eight, and we’ve already found what this is. This is negative two. And so 𝑓 prime of eight is a third times one over negative two squared, which is a third times a fourth as negative two squared is four. And a third times a fourth is a twelfth. Great! So let’s substitute that in. Everything else is just algebra. With the substitutions, we get 𝑓 of π‘₯ being approximately a twelfth π‘₯ minus negative eight plus negative two when π‘₯ is approximately negative eight.

Subtracting a negative eight is the same as adding eight. And adding negative two is the same as subtracting two. We expand the parenthesis and notice that the fraction eight over 12 can be simplified to two over three. We combine our constant terms using the fact that two is six over three to get our final answer 𝑓 of π‘₯ is approximately equal to a twelfth π‘₯ minus four over three when π‘₯ is approximately negative eight.

Let’s see how good our linear approximation is by approximating 𝑓 of negative seven. We need to calculate a twelfth of negative seven minus four over three. Calculating this gives negative 1.916 recurring, and our calculator gives us a value of negative 1.9129 dot dot dot. Our linear approximation is therefore pretty good. Of course, as we saw from the graph, this approximation is only good when π‘₯ is near negative eight. We can try our linear approximation with π‘₯ equal to one.

Our approximation gives the cube root of one as negative 1.25, which is well off from the true value of one. We can take what is currently the top line of working, 𝑓 of π‘₯ is approximately 𝑓 prime of negative eight times π‘₯ minus negative eight plus 𝑓 of negative eight when π‘₯ is approximately negative eight, and generalize it. Replacing negative eight by π‘Ž, we get 𝑓 of π‘₯ is approximately equal to 𝑓 prime of π‘Ž times π‘₯ minus π‘Ž plus 𝑓 of π‘Ž. And this is the linear approximation of 𝑓 at π‘Ž.

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