Home > Calculus > Identifying Stationary Points (Critical Points) for a Function

How do you find the critical points for #f(x)= (2x^2+5x+5)/(x+1)#?

Answer 1

Elijah Abram

#(-2, -3), (0, 5)#

The derivative is either 0 or undefined:

#f'(x) = ((4x + 5)(x + 1) - 1(2x^2 + 5x + 5))/(x + 1)^2# #f'(x) = (2x^2 + 4x)/(x + 1)^2#

#f'(x) = 0# #(2x^2 + 4x)/(x + 1)^2=0# #2x(x+2) = 0# #x = -2, x = 0#

#f(-2) = -3# #f(0) = 5#

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Answer 2

Evelyn Agard

To find the critical points of ( f(x) = \frac{2x^2 + 5x + 5}{x + 1} ), you need to follow these steps:

Find the derivative of the function using the quotient rule.
Set the derivative equal to zero and solve for ( x ).
Check the second derivative test to confirm whether the critical points are maximum, minimum, or inflection points.

First, find the derivative:

[ f'(x) = \frac{(2x+5)(x+1) - (2x^2+5x+5)}{(x+1)^2} ]

[ f'(x) = \frac{2x^2 + 2x + 5x + 5 - 2x^2 - 5x - 5}{(x+1)^2} ]

[ f'(x) = \frac{-3}{(x+1)^2} ]

Set ( f'(x) ) equal to zero:

[ -3 = 0 ]

Since this equation has no solutions, there are no critical points.

Therefore, the function ( f(x) = \frac{2x^2 + 5x + 5}{x + 1} ) has no critical points.

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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.