A function where the derivative is given by f ' (x) =3x^2+4x-7 has a local maximum at x= ???
2007-12-23 15:42:58 · 3 answers · asked by atlien2008 1 in Science & Mathematics ➔ Mathematics
where the derivative is equal to 0. So set f' to 0 and solve for x.
Since you are in AP Cal, i don't need to show the simple algebra on how to solve for it. :)
Edit: the person above is crazy, the derivative is quadratic, so the function is cubic, which means it must have one local maximum. This can be proven by finding the anti-derivative, ((3x^3)/3)+(4x^2/2)-7x+C. You will learn this later on :)
Why use the Quadratic Formula, when it factors perfectly.
3x^2+4x-7=0
(3x+7)(x-1)=0
2007-12-23 15:47:15 · answer #1 · answered by sirdumbalot8 3 · 0⤊ 0⤋
To find maxima or minima, you need to first find critical points---that is, the points at which the derivative is 0. To do this, set your derivative---in this case, f'(x)=3*x^2+4*x-7---to 0. Then solve for x. So we have:
f'(x)=3*x^2+4*x-7=0.
Use the quadratic formula to solve for x.
Recall that the quadratic formula for a*x^2 + bx + c = 0 is defined as:
x = [-b +/- sqrt(b^2 - 4*a*c)]/(2*a)
Can you take it from here?
2007-12-23 23:50:46 · answer #2 · answered by A.K. 1 · 0⤊ 1⤋
It doesn't. If the quadratic term is positive, it has one and only one absolute minimum.
2007-12-23 23:46:53 · answer #3 · answered by cattbarf 7 · 0⤊ 1⤋