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

Oct 22, 2017

Let's see.

Explanation:

Let the function given be $y$ such that $\rightarrow$

$y = f \left(x\right) = - {x}^{2} + 2 x + 3$

Now differentiating w.r.t $x$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x + 2$

Now the second order derivative is :

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - 2$

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 $f \left(- 2\right)$.

Hope it Helps:)

Oct 22, 2017

Let's see.

Explanation:

Let the function given be $y$ such that $\rightarrow$

$y = f \left(x\right) = - {x}^{2} + 2 x + 3$

Now differentiating w.r.t $x$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x + 2$

Now the second order derivative is :

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - 2$

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 $f \left(- 2\right)$.

Hope it Helps:)