Question

Need help ASAP- calculus question

Answer 1

i know its not D, stuck between B and C i think

Answer 2

The Taylor Expansion of a function \(f(x)\) about \(x\), for \(|\varepsilon| << 1\), is:

\(f(x + \varepsilon) = f(x) + \varepsilon f'(x) + \dfrac{\varepsilon^2}{2!} f''(x) + \dfrac{\varepsilon^3}{3!} f'''(x) + \cdots \)

But the **local linear approximation**, we can call it \(L(x)\), is limited to the 1st derivative and is:

\(L(x) = f(x) + \varepsilon f'(x) \)


We can then define the difference, \(\Delta\), as:

\(\Delta = f(x + \varepsilon) - L(x) \)

\(=\dfrac{\varepsilon^2}{2!} f''(x) + \dfrac{\varepsilon^3}{3!} f''(x) + \cdots\)


For \(|\varepsilon| << 1\), the second derivative is the dominant term. And it is squared, as in \(\varepsilon^2\), so the sign of \(\Delta\) turns totally on the sign of the second derivative.

So \(\Delta > 0\) if \(f''(x) > 0\)

You need to go through this yourself BTW as I just made it up and typed it in. It's not an easy question to face cold, IMHO, FWIW.

Answer 3

Shoulda said for clarity, same applies for \(f(x - \varepsilon)\)

The qu isn't particular about which way you look on the curve.

:)

Answer 4

@sillybilly123 so what would they be please?

Answer 5

@sillybilly123 its A then right?

Answer 6

please someone? @Yinshy

Answer 7

@steve816

Answer 8

yep

this approach suggests A

Answer 9

can you help me with a few more please??? @sillybilly123

Answer 10

ya shoulda posted the qu, then coulda helped instead of saying "i will help if i can" !!

:)