Video: Using the Maclaurin Series to Find the Equation of a Tangent

If the Maclaurin series of the function 𝑓 is 𝑓(π‘₯) = 2 βˆ’ (1/6 π‘₯) + (5/24 π‘₯Β²) βˆ’ (7/60 π‘₯Β³) + (3/80 π‘₯⁴) + β‹…β‹…β‹…, find the equation of the tangent to the curve of 𝑓 at π‘₯ = 0.

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Video Transcript

If the Maclaurin series of the function 𝑓 is 𝑓 of π‘₯ is equal to two minus one over six multiplied by π‘₯ plus five divided by 24 multiplied by π‘₯ squared minus seven divided by 60 multiplied by π‘₯ cubed plus three divided by 80 multiplied by π‘₯ to the fourth power and so on. Find the equation of the tangent to the curve of 𝑓 at π‘₯ is equal to zero.

The question gives us the Maclaurin series for a function 𝑓 and asks us to find the equation of the tangent to the curve of 𝑓 when π‘₯ is equal to zero. We recall that a straight line will have equation 𝑦 is equal to π‘šπ‘₯ plus 𝑐 where π‘š is the slope of our line and 𝑐 is the 𝑦-intercept. We also recall that our function 𝑓 prime of π‘₯ gives us the slope of the tangent to the curve of 𝑓 at π‘₯. And since the question asks us to find the tangent when π‘₯ is equal to zero, we can conclude that the slope of our tangent is given by 𝑓 prime of zero.

Now, we recall that the Maclaurin series of a function is equal to the Taylor series expansion about zero. Namely, this gives us that 𝑓 of π‘₯ is equal to 𝑓 evaluated at zero plus the first derivative of 𝑓 evaluated at zero multiplied by π‘₯ plus the second derivative of 𝑓 evaluated at zero divided by two factorial multiplied by π‘₯ squared plus the third derivative of 𝑓 evaluated at zero divided by three factorial multiplied by π‘₯ cubed. And if we keep going and going, then we get the Maclaurin series for our function 𝑓.

From the question we’re given the Maclaurin series of our function 𝑓 is equal to two plus negative one-sixth multiplied by π‘₯ plus five over 24 multiplied by π‘₯ squared plus negative seven divided by 60 multiplied by π‘₯ cubed. And this keeps going and going. We can see from the definition of our Maclaurin series that the coefficient of π‘₯ is equal to 𝑓 prime of zero. Therefore, we must have that the first derivative of 𝑓 evaluated at zero is equal to negative one-sixth. So we have the slope of our tangent is negative one-sixth giving us the equation 𝑦 is equal to negative one-sixth multiplied by π‘₯ plus some constant 𝑐.

One way to find the value of 𝑐 is to find the π‘₯- and 𝑦-coordinates of a point on our tangent line. And one way to do this is to notice that since we’re finding the tangent line to the curve of 𝑓 when π‘₯ is equal to zero, we must have that the tangent line and the curve of 𝑓 meet when π‘₯ is equal to zero. Therefore, we must have that when π‘₯ is equal to zero, our 𝑦-value is equal to the function 𝑓 evaluated at zero. Now, we just need to find the value of the function 𝑓 evaluated at zero. And we can do this by looking on our Maclaurin series. We can see that the first term in our series is equal to the function 𝑓 evaluated at zero. So we must have that 𝑓 zero is equal to two.

So what we have just shown is that the 0.2 is on our tangent line. Therefore, if we substitute the value of π‘₯ equals zero and the value of 𝑦 equals two into the equation for our line, we get two is equal to negative one-sixth multiplied by zero plus 𝑐. And since negative one-sixth multiplied by zero is just equal to zero, we must have that 𝑐 is equal to two. So what we’ve shown is that for the Maclaurin series 𝑓 of π‘₯ given to us in the question, the equation of the tangent to the curve 𝑓 at π‘₯ is equal to zero is 𝑦 is equal to negative one-sixth π‘₯ plus two.

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