Question

Let fx=ln(2x-3). Find a quadratic polynomial p(x) so that p2=f2, p'2=f'2 and p''2=f''2.

p(x)=

Answer 1

start with p(x) = ax^2+bx+c ... differentiate and you'll have a system (3 eq)

Answer 2

Let \(p(x) = ax^2 + bx + c\) and f(x) is defined as \( \ln {2x-3}\)
Condition 1. \(p(2)=f(2)\)
\[4a +2b+c = \ln1 \implies 4a + 2b+c = 0\]
Condition 2. \(p'(2) = f'(2)\)
\[2a\times 2 + b = \frac{2}{2 \times 2 - 3} \implies 4a + b = 2\]
Condition 3. \(p"(2) = f"(2)\)
\[2a = 2\frac{-2}{(2*2-3)^2}\implies a = -2 \]

I didn't solve the question on paper, so chances are I might have done something wrong somewhere. If it's homework question then don't copy it from my answer. Get the Idea of what I did and then try to solve it again.

Answer 3

ok thanks

Answer 4

And yeah you get 3 equations after applying the conditions, solve the three equations, get the values of the co-efficient of the quadratic polynomial.

Answer 5

so it will be x^2+2x-2

Answer 6

According to my solution, a=-2, b=10 and c= -12. I think my solution might be wrong, let me try it on the paper.

Answer 7

its correct. thanks ishaan94

Answer 8

Yeah, it's correct.