How do you find the local extrema for #f(x) = 2-2x^2# on domain #-1 <= x <= 1#?

Answer 1

William Abdalla

It has maximum for #x=0# and value #f(0)=2# which is a absolute maximum because

#f(x)<=f(0)# for all x in #[-1,1]#

Also using derivatives we find the roots of first derivative which is

#(df(x))/dx=(d(2-2x^2))/dx=-4x#

But #(df)/dx=0=>-4x=0=>x=0#

Hence point #x=0# is a critical value of #f(x)#.Above we show that this a maximum.

The graph of #f(x)# is as follows

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Answer 2

Charles Arocha

To find the local extrema of $f(x) = 2 - 2x^2$ on the domain $-1 \leq x \leq 1$ , we first need to find the critical points by taking the derivative of the function and setting it equal to zero. Then, we check the second derivative to determine the nature of the critical points.

First, find the derivative of $f(x)$ : $f'(x) = -4x$

Set $f'(x) = 0$ to find critical points: $-4x = 0$ $x = 0$

Now, since the domain is restricted to $-1 \leq x \leq 1$ , we evaluate the function at the critical point and at the endpoints of the domain: $f(-1) = 2 - 2(-1)^2 = 0$ $f(0) = 2 - 2(0)^2 = 2$ $f(1) = 2 - 2(1)^2 = 0$

Since the function goes from positive (at $x = 0$ ) to negative (at $x = -1$ ) to positive (at $x = 1$ ), $f(x)$ has a local maximum at $x = 0$ and local minima at $x = -1$ and $x = 1$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the local extrema for #f(x) = 2-2x^2# on domain #-1 &lt;= x &lt;= 1#?

How do you find the local extrema for #f(x) = 2-2x^2# on domain #-1 <= x <= 1#?