Finding local maximums and minimums

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Homework Statement


Find a cubic function g(x)=ax^3 +bx^2 +cx +d that has a local maximum value of 2 at -9, and a local minimum value of -7 at 8.


Homework Equations





The Attempt at a Solution


I thought i would find the derivative and set it equal to zero, but i do not know what to do from there
 
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Well, to find the values of a,b,c,d you'll need some constraints on those numbers

1) You know that when you plug 2 in your poly, you should get -9. When you plug -7 in, you would get 8. These deliver two equations.

2) When you derive your function, you know that when you plug 2 in, you get 0. When you plug -7 in, you would get 0 to. These deliver another two equations.

So you have 4 equations. You should be able to solve this system without problems...
 
So plug those numbers in for x? correct?
 
yes!
 
Awesome. i don't know why i didnt get that
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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