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

If the Maclaurin series of the function 𝑓 is 𝑓(𝑥) = 2 − (1/6 𝑥) + (5/24 𝑥²) − (7/60 𝑥³) + (3/80 𝑥⁴) + ⋅⋅⋅, find the equation of the tangent to the curve of 𝑓 at 𝑥 = 0.

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### Video Transcript

If the Maclaurin series of the function 𝑓 is 𝑓 of 𝑥 is equal to two minus one over six multiplied by 𝑥 plus five divided by 24 multiplied by 𝑥 squared minus seven divided by 60 multiplied by 𝑥 cubed plus three divided by 80 multiplied by 𝑥 to the fourth power and so on. Find the equation of the tangent to the curve of 𝑓 at 𝑥 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 𝑥 is equal to zero. We recall that a straight line will have equation 𝑦 is equal to 𝑚𝑥 plus 𝑐 where 𝑚 is the slope of our line and 𝑐 is the 𝑦-intercept. We also recall that our function 𝑓 prime of 𝑥 gives us the slope of the tangent to the curve of 𝑓 at 𝑥. And since the question asks us to find the tangent when 𝑥 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 𝑥 is equal to 𝑓 evaluated at zero plus the first derivative of 𝑓 evaluated at zero multiplied by 𝑥 plus the second derivative of 𝑓 evaluated at zero divided by two factorial multiplied by 𝑥 squared plus the third derivative of 𝑓 evaluated at zero divided by three factorial multiplied by 𝑥 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 𝑥 plus five over 24 multiplied by 𝑥 squared plus negative seven divided by 60 multiplied by 𝑥 cubed. And this keeps going and going. We can see from the definition of our Maclaurin series that the coefficient of 𝑥 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 𝑥 plus some constant 𝑐.

One way to find the value of 𝑐 is to find the 𝑥- 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 𝑥 is equal to zero, we must have that the tangent line and the curve of 𝑓 meet when 𝑥 is equal to zero. Therefore, we must have that when 𝑥 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 𝑥 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 𝑥 given to us in the question, the equation of the tangent to the curve 𝑓 at 𝑥 is equal to zero is 𝑦 is equal to negative one-sixth 𝑥 plus two.