Page:Calculus Made Easy.pdf/135

 into a downward or negative slope, so that in this case the “slope of the slope” $$\dfrac{d^2y}{dx^2}$$ is negative.

Go back now to the examples of the last chapter and verify in this way the conclusions arrived at as to whether in any particular case there is a maximum or a minimum. You will find below a few worked out examples.

Go back now to the examples of the last chapter and verify in this way the conclusions arrived at as to whether in any particular case there is a maximum or a minimum. You will find below a few worked out examples.

(1) Find the maximum or minimum of

and ascertain if it be a maximum or a minimum in each case.

(a) $$\dfrac{dy}{dx} = 8x-9=0$$; $$x=1\tfrac{1}{8}$$; and $$y = -11.065$$.


 * $$\dfrac{d^2y}{dx^2} = 8$$; it is $$+$$; hence it is a minimum.

(b) $$\dfrac{dy}{dx}=9-8x=0$$; $$x=1\tfrac{1}{8}$$; and $$y=+11.065$$.


 * $$\dfrac{d^2y}{dx^2} = -8$$; it is $$-$$; hence it is a maximum.

(2) Find the maxima and minima of the function $$y=x^3-3x+16$$.

hence $$x=1$$ corresponds to a minimum $$y=14$$. For $$x=-1$$ it is $$-$$; hence $$x=-1$$ corresponds to a maximum $$y=+18$$.