How do you find the critical numbers of #f(x)=(x^2+6x-7)^2#?

Question

Charles Anctil · Accepted Answer

#x=-7, x=1# and #x=-3# The critical points of a function are where the function's derivative is either undefined or #0#. Let's first start by computing #f'(x)#. I will not expand the parenthesis (because I'll have to do less factoring later) and instead use the chain rule. If we let #u=x^2+6x-8#, we get: #d/dx((x^2+6x-7)^2)=d/(du)(u^2)d/dx(x^2+6x-7)# #2u(2x+6)=2(x^2+6x-7)(2x+6)# Now we set this expression equal to #0#. #2(x^2+6x-7)(2x+6)=0# Factoring #x^2+6x-7# gives: #2(x+7)(x-1)(2x+6)=0# This tells us that #x=-7, x=1# and #x=-3# all are solutions and therefor critical points. This function is never undefined, so these are also the only critical points.

Isabella Ackerman · Answer

undefined To find the critical numbers of $ f(x) = (x^2 + 6x - 7)^2 $, you first find the derivative of the function and then solve for values of $ x $ where the derivative equals zero or is undefined.

1. Find the derivative of $ f(x) $ using the chain rule and power rule:
$$ f'(x) = 2(x^2 + 6x - 7)(2x + 6) $$

2. Set the derivative equal to zero and solve for $ x $:
$$ 2(x^2 + 6x - 7)(2x + 6) = 0 $$
$$ (x^2 + 6x - 7)(2x + 6) = 0 $$

3. Solve each factor for zero:
$$ x^2 + 6x - 7 = 0 $$ and $$ 2x + 6 = 0 $$

4. Solve the quadratic equation $ x^2 + 6x - 7 = 0 $ using the quadratic formula or factoring.

5. Solve $ 2x + 6 = 0 $ for $ x $.

6. The values of $ x $ obtained from steps 4 and 5 are the critical numbers of the function.