Find the values of x where f has a local max and min

Question

f(x)= ln(2x+4)+x^2

Answer 1

f(x) = ln(2x+4) + x²  we can't take logs of nonpositive numbers, so 2x + 4 > 0.  Therefore, the domain of f is x > -2.

 

f'(x) = 2/(2x+4) + 2x = [2 + 2x(2x+4)]/(2x+4)

 

                                        = [4x² + 8x + 2]/(2x+4)

 

f'(x) = 0 when 4x² + 8x + 2 = 0

 

Divide both sides by 2:  2x² + 4x + 1 = 0

 

                                         x = [-4 ±√(16 - 4(2)(1))]/4

 

                                             =[-4 ± √8]/4 ≈ -0.29, -1.7

 

(    +    l          -         l       +     sign f'

-2     -1.7           -0.29

 

The derivative changes sign from + to - at -1.7 and changes sign from - to + at -0.29, so there is a:

 

relative max at x = [-4 - √8]/4 ≈ -1.7

 

relative min at x = [-4 + √8]/4 ≈ -0.29

Answer 2

f'(x) = 2/(2x+4) + 2x. Setting equal to zero leads to insoluble quadratic. Hm mm.