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Find the linear approximation of the function π of π₯ equals the cube root of π₯ at π₯ equals negative eight.
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We have a function, the cube root function, and we know some values that that function takes.
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The cube root of one is one because one cubed is one.
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The cube root of eight is two because two cubed is eight.
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And the cube root of negative one twenty-seventh is negative a third because negative a third cubed is negative one twenty-seventh.
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But how do you find the cube root of negative seven?
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There doesnβt seem to be an easy way without using a calculator.
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But how does your calculator find the cube root of negative seven.
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One way is to find a linear approximation of our function, the cube root function.
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We approximate π of π₯ by the linear function which takes π₯ to ππ₯ plus π for some numbers π and π.
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We can then use this linear function to find an approximate value for the cube root of negative seven.
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But how do we find a linear approximation of our function?
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Letβs take a look at its graph.
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Here is a graph of our cube root function.
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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 π of π₯ equals the cube root of π₯ for values of π₯ near negative eight.
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The graph of this linear function will be a straight line.
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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.
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Can we think of what straight line this might be?
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A good choice is the tangent to our graph at π₯ equals negative eight.
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The exact value of π of negative seven is the π¦-coordinate of the point on the graph of π of π₯ with π₯-coordinate negative seven.
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And we can see from the graph that this is very close to the π¦-coordinate of the point on the tangent at π₯ equals negative seven.
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We are, therefore, looking for the equation of the tangent.
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How do we find the equation of the tangent?
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Well we can differentiate to find the slope of the tangent.
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This slope is π prime of negative eight.
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And the tangent passes through the point negative eight π of negative eight, which is where the tangent touches the graph.
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Itβs therefore easy to find the equation of this line in point-slope form.
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The equation of the line with slope π which passes through the point π₯ naught, π¦ naught is π¦ equals π times π₯ minus π₯ naught plus π¦ naught.
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And so as the slope π is π prime of negative eight, π₯ naught is negative eight and π¦ naught is π of negative eight.
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We find that π¦ is π prime of negative eight times π₯ minus negative eight plus π of negative eight.
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And remember, weβre using the equation of this tangent to approximate π of π₯ near π₯ equals negative eight.
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Okay letβs now clear some room.
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We just need to find the values of π prime of negative eight and π of negative eight now.
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Letβs start with π of negative eight.
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This is the cube root of negative eight.
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And this is equal to negative two, as negative two cubed is negative eight.
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π prime of negative eight is slightly harder to find.
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We have to differentiate π.
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So we have to differentiate the cube root of π₯.
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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 π with respect to π₯ which works not just when π is a whole number or integer but for any real value of π.
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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.
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π 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.
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Using what we know about negative exponents, using more exponent laws, this is a third times one over negative eight to the third squared.
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Negative eight to the power of a third is the cube root of negative eight, and weβve already found what this is.
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This is negative two.
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And so π 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.
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And a third times a fourth is a twelfth.
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Great!
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So letβs substitute that in.
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Everything else is just algebra.
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With the substitutions, we get π of π₯ being approximately a twelfth π₯ minus negative eight plus negative two when π₯ is approximately negative eight.
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Subtracting a negative eight is the same as adding eight.
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And adding negative two is the same as subtracting two.
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We expand the parenthesis and notice that the fraction eight over 12 can be simplified to two over three.
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We combine our constant terms using the fact that two is six over three to get our final answer π of π₯ is approximately equal to a twelfth π₯ minus four over three when π₯ is approximately negative eight.
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Letβs see how good our linear approximation is by approximating π of negative seven.
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We need to calculate a twelfth of negative seven minus four over three.
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Calculating this gives negative 1.916 recurring, and our calculator gives us a value of negative 1.9129 dot dot dot.
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Our linear approximation is therefore pretty good.
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Of course, as we saw from the graph, this approximation is only good when π₯ is near negative eight.
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We can try our linear approximation with π₯ equal to one.
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Our approximation gives the cube root of one as negative 1.25, which is well off from the true value of one.
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We can take what is currently the top line of working, π of π₯ is approximately π prime of negative eight times π₯ minus negative eight plus π of negative eight when π₯ is approximately negative eight, and generalize it.
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Replacing negative eight by π, we get π of π₯ is approximately equal to π prime of π times π₯ minus π plus π of π.
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And this is the linear approximation of π at π.