Home > Calculus > Identifying Turning Points (Local Extrema) for a Function

What are the local extrema, if any, of #f (x) =x^2-2x+4#?

Answer 1

Charles Arocha

#f(1) = 3# is a local minimum.

#f'(x) = 2x-2#

Critical number #x=1#.

#f'(x) < 0# for #x < 1# and #f'(x) > 0# for #x > 1#, so #f(1) = 3# is a local minimum.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Elijah Abram

To find the local extrema of the function ( f(x) = x^2 - 2x + 4 ), we first need to find its derivative. The derivative of ( f(x) ) with respect to ( x ) is ( f'(x) = 2x - 2 ).

To find the critical points, we set the derivative equal to zero and solve for ( x ): [ 2x - 2 = 0 ] [ 2x = 2 ] [ x = 1 ]

So, the critical point is ( x = 1 ).

To determine the nature of the extrema at this critical point, we can use the second derivative test. The second derivative of ( f(x) ) is ( f''(x) = 2 ), which is positive for all ( x ). Since the second derivative is positive, the function is concave up everywhere, including at the critical point.

Therefore, at ( x = 1 ), ( f(x) ) has a local minimum.

So, the local extrema of ( f(x) = x^2 - 2x + 4 ) is a local minimum at ( x = 1 ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.