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**phoenix_7** 1. If the graph of f(x)=2x^2+k/x has a point of inflection at x=-1, what is the value of k?

**Points of inflection happen when f'(x) = 0... f'(x) = 4x-k/x^2 = 0.**

x=-1

Hence

4(-1)-k/(-1)^2=0... solve for k!

2. If f(x)=3x^2-8x^-2, then lim f(2+h)-f(2)/h =?

h->0

** Well you could first realise that lim(h->0) (f(x+h) - f(x))/h is equal to the derivative of f(x). And hence lim(h->0) (f(2+h) - f(2))/h is the derivative of f(x) at the value to. f'(x) = 6x-8 = lim(h->0) (f(x+h) - f(x))/h. Hence f'(2) = lim(h->0) (f(2+h) - f(2))/h = 6(2)-8**

3. For what values of x is the graph of y=2/(4-x) concave downward?

** 2nd derivative test helps you out here!**

4. If the graph of f(x)=Ax^2+Bx+C where A,B, and C are constants, passes through the point (-1,-15) and attains a relative maximum at the point (2,3), then what are the values of A,B, and C?

**It goes through the point (-1,-15), so plug those coordinates into the equation! -15 = A(-1)^2+B(-1)+C**

It also goes through (2,3), since that's where the maximum is. Plug those coordinates in too.

3 = A(2)^2+B(2)+C

And there's a maximum at the x coordinate 2. Maximum is when the derivative is zero

f'(x) = 2Ax+B = 0

and x = 2 is a solution, since the maximum is at (2,3).

4A+B = 0. Hence you can get B in terms of A (B = -4A), and plug it into the two equations above, then you'll have 2 equations in two unknowns which are solveable!

I need to know that answers ASAP because i have a final next week and i need to know this. thanks a lot.