189k views
5 votes
Stationary points. Help ASAP please. Thanks

Stationary points. Help ASAP please. Thanks-example-1
User Pzirkind
by
8.0k points

1 Answer

6 votes

You can use the power rule for derivatives for each problem (though you could also use the product rule for the fourth curve).

y = x ² + 6x - 1 ==> dy/dx = 2x + 6

y = x ² - 5x + 1 ==> dy/dx = 2x - 5

y = 2 - 4x - x ² ==> dy/dx = -4 - 2x

y = (1 + x) (7 - x) = 7 + 6x - x ² ==> dy/dx = 6 - 2x

--

or, using the product rule,

dy/dx = (1 + x) (-1) + 1 (7 - x) = -1 - x + 7 - x = 6 - 2x

--

Now, stationary points occur where the derivative is zero. We have

2x + 6 = 0 ==> x = -3

2x - 5 = 0 ==> x = 5/2

-4 - 2x = 0 ==> x = -2

6 - 2x = 0 ==> x = 3

User Ddotsenko
by
7.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories