What are the extrema of #f(x)=3+ 2x -x^2#?

2 Answers
Oct 22, 2017

Answer:

Let's see.

Explanation:

Let the function given be #y# such that #rarr#

#y=f(x)=-x^2+2x+3#

Now differentiating w.r.t #x#:

#dy/dx=-2x+2#

Now the second order derivative is :

#(d^2y)/dx^2=-2#

Now, the second order derivative is negative.

Hence, the function only has an extrema & no minima.

Therefore the point of maxima is #-2#.

The maximum value of the function is #f(-2)#.

Hope it Helps:)

Oct 22, 2017

Answer:

Let's see.

Explanation:

Let the function given be #y# such that #rarr#

#y=f(x)=-x^2+2x+3#

Now differentiating w.r.t #x#:

#dy/dx=-2x+2#

Now the second order derivative is :

#(d^2y)/dx^2=-2#

Now, the second order derivative is negative.

Hence, the function only has an extrema & no minima.

Therefore the point of maxima is #-2#.

The maximum value of the function is #f(-2)#.

Hope it Helps:)