find the max/min of a function?

Question

im supposed to find either the max/min of this function.
help?
f(x)= -x^2 - 2x

Answer 1

The max/min can be defined as the point on the parabola where the slope is zero. By definition, the derivative of a function is a function of the original function's slope. Thus, if we take the derivative and find out where it is zero, we can find the min/max.



f(x) = -x^2 - 2x



f ' (x) = -2x - 2



0 = -2x - 2



x = -1



When x = -1, there is a min/max. Let's solve for y:



f(x) = -(-1)^2 - 2(-1) = -1 + 2 = 1



The point of the min/max is (-1,1).

Answer 2

Take the derivative and calculate the for 0.



f(x)= -x^2 - 2x



f'(x) = -2x - 2 <--- this is the first derivative, now solve for 0



-2x - 2 = 0

-2x = 2

-x = 1

x = -1



Now plug -1 back into the original equation to get your y-value.



f(-1) = -(-1)^2 - 2(-1)

f(-1) = -1 + 2

f(-1) = 1



The point is (-1, 1)

This is your min/max point

Answer 3

The Max is



1 at x= -1



The Min is



- Inf at x = - Inf

Answer 4

The max is ( - 1, 1).



Download Graph 4.4 from www.padowan.dk for free.



On "Insert function", type (- x^2 - 2*x), then "OK".

Answer 5

y=48x^2-24x+c might have 2 intercepts whilst Determinant >0 i.e 576-4*40 8*c>0 576-192c>0 192c<576 c<3 the utmost fee of a quadratic fee whilst a>0 is infinity and minimum fee is -D/4a the place D is discriminant