# need answers to these 4 questions soon!!

• Jan 5th 2009, 04:04 PM
phoenix_7
need answers to these 4 questions soon!!
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?

2. If f(x)=3x^2-8x^-2, then lim f(2+h)-f(2)/h =?
h->0

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

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?

I need to know that answers ASAP because i have a final next week and i need to know this. thanks a lot.
• Jan 5th 2009, 04:14 PM
Mush
Quote:

Originally Posted by 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.

Mush
• Jan 5th 2009, 04:23 PM
phoenix_7
thanks, And i think you misread #2. the -2 is raised to the 8. and, i still didn't get the answer to #1.
• Jan 5th 2009, 05:45 PM
mr fantastic
Quote:

Originally Posted by 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?

[snip]

At a point of inflection $\displaystyle f''(x) = 0$ (note that the converse is not always true).

Assuming by f(x)=2x^2+k/x you mean f(x)=2x^2 + (k/x), that is, $\displaystyle f(x) = 2x^2 + \frac{k}{x}$ then you need to substitute $\displaystyle x = 1$ into $\displaystyle 0 = 4 + \frac{2k}{x^3}$ and then solve for $\displaystyle k$.

Quote:

Originally Posted by mush
Quote:

Originally Posted by 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!
[snip]

Mush

Uh-uh. See above.
• Jan 5th 2009, 05:49 PM
mr fantastic
Quote:

Originally Posted by phoenix_7
thanks, And i think you misread #2. the -2 is raised to the 8. [snip]

Regardless of the misreading, the approach suggested by Mush is valid. Try applying it. I assume you can differentiate f(x) ....?