Question. Set f(x)=ln(x).a. Find the 1st and 2nd Taylor Polynomials at 2 of f(x).b. Complete the following steps to determine how well the 1st and 2nd Taylor Polynomials at 2 of f(x) approximate f(x) on the interval 2,2.5? We want to find the largest possible value of the error term. Recall that the error terms are1st Taylor Polynomialf(2)(c)2(x-2)22nd Talyor Polynomialf(3)(c)6(x-2)3So, compute f(2)(x)=, and f(3)(x)= For x2, the graph of |f(2)(x)| and |f(3)(x)| aregraph the functions.On 2,2.5, the graph of |x-2|2 and |x-3|3 aregraph the functions.Therefore, the largest values of |(x-2)2| and |(x-2)3| on 2,2.5 are|(x-2)2|,|(x-2)3|We now conclude that on the interval 2,2.5,1st Taylor Polynomial2nd Talyor Polynomial|f(2)(c)2(x-2)2|,|f(3)(c)6(x-2)3|Conclusion: On the interval 2,2.5,i. The 1st Taylor polynomia approximates f(x)=ln(x) to withinandii. the 2nd Taylor polynomialapproximates f(x)=ln(x) to withinc. Complete the following steps to find the time interval for which the 1st and 2nd Taylor Polynomials at 2 approximate f(x)=ln(x) to within 12000 for x2. Recall that the error terms are1st Taylor Polynomialf(2)(c)2(x-2)22nd Talyor Polynomialf(3)(c)6(x-2)3The graphs of f(2)(x) and f(3)(x) in Part (b) imply that for x2,1st Taylor Polynomial|f(2)(c)2|2nd Talyor Polynomial|f(3)(c)6|and the inequalities we want to solve are1st Taylor Polynomial|x-2|2120002nd Talyor Polynomial*|x-2|312000Solve the inequalitiesConclusion:i. On the interval the 1st Taylor Polynomial at 2 approximates f(x)=ln(x) to within 12000 andii. On the intervalthe 2nd Taylor Polynomial at 2 approximates f(x)=ln(x) to within 12000.