Question

calc

Answer 1

yes , but keep in ur mind the tangents would be negative

Answer 2

I would look at the second derivative.

Answer 3

but how did u get -4 ? i think its not -4 thought let me check

Answer 4

oh my bad \(\phi\) :P

Answer 5

\(\dfrac{d}{dx}\dfrac{-3}{x+4}=\dfrac{3}{(x+4)^2}\)

So \(\dfrac{d^2}{dx^2}\dfrac{-3}{x+4}=\dfrac{-6}{(x+4)^3}\)

For this to be negative (f(x) concaves down when f''(x) < 0), x+4 have to be positive, so you have \(x+4>0\)

\(\boxed{x>-4}\)

Answer 6

@geerky42 wouldnt it be less than?

Answer 7

concave up "like a cup"
concave down , "like a frown"

Answer 8

No, we want f''(x) to be negative, and numerator is already negative, so we want denominator to stay positive.

So you have \(x+4>0\)

Subtract both sides by 4 and you have \(x > -4\)

Answer 9

and we know x^2 (a parabola) has second derivative = +2 i.e. plus second derivative means concave up

Answer 10

ohh

Answer 11

I had made a little reference guide last year to questions relating to the first and second derivative tests for inc/dec functions and concavity.

Hope this helps!