Question

mean value theorem help, - no -

Answer 1

having a hard time working this one backwards

Answer 2

yes ;)

Answer 3

Doing the second one:

You need \(f'(x) = 6x^2-2x\) and \(f(2) = 13\)

\[f(x) = 2x^3-x^2+C\]which gives us \[f'(x) = 3*2x^{3-1}-2x^{2-1} = 6x^2-2x\]\[f(2)=13\]\[2(2)^3-(2)^2+c = 13\]\[2*8-4+c=13\]\[16-4+c=13\]\[c=1\]so\[f(x) = 2x^3-x^2+1\]Checking our work:
\[2(2)^3-(2)^2+1 = 13\]\[16-4+1=13\]passes through the specified point
\[\frac{d}{dx}[2x^3-x^2+1] = 6x^2-2x\]has the right derivative.

Answer 4

For the first one, we want to find functions that have the derivative \[10x+6\]
Such a function is going to have \(ax^2 + bx+C\) as its general form. If we take the derivative of \(ax^2+bx + C\) we get \(2ax^{2-1} + bx^{1-1} = 2ax+b\)

If we set that equal to \(10x+6\)\[2ax+b=10x+6\]It doesn't take much to figure out the values of \(a,b\)

Answer 5

To do the second part, we take our new \(f(x) = ax^2+bx+C\) and plug in the known values of \((1,21)\):
\[21 = a(1)^2 + b(1) + C\]\[21 = a+b + C\]and we already know the values of \(a,b\) from the previous response, so it's just simple arithmetic to find C.

Any questions?

Answer 6

got it! Youre the best! I wish I could show my titties but the admins said I'd be banned if i did so.