Let f be a differentiable function such that f(3) = 2.345 and f' (x) = ln (x^2+ 1). What is the value of f (5)?
A) 3.301
B) 5.631
С) 6.950
D) 7.976

Here we want to solve the differential equation, by finding the Taylor expansion of the function and evaluation we will find that f(5) = 6.950, so the correct option is C.

Taylor series the equation:

Here we must start by doing a Taylor expansion of the function, remember that the Taylor expansion is given by:

f(x) = f(a) + f'(a)*(x - a) + ...

The series has more terms, but we can work with two to get an estimation, if we use a = 3 we get:

f(x) = f(3) + ln(3^2 + 1)*(x - 3) = 2.345 + ln(10)*(x - 3)

Now we evaluate this in x = 5 to get:

f(5) = 2.345 + ln(10)*(5 - 3) = 6.950

So the correct option is C.

If you want to learn more about Taylor series, you can read:

https://brainly.com/question/9211177

Let f be a differentiable function such that f(3) = 2.345 and f' (x) = ln (x^2+ 1). What is the value of f (5)? A) 3.301 B) 5.631 С) 6.950 D) 7.976

Respuesta :

Taylor series the equation:

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Let f be a differentiable function such that f(3) = 2.345 and f' (x) = ln (x^2+ 1). What is the value of f (5)?
A) 3.301
B) 5.631
С) 6.950
D) 7.976