need help Let f(x) = k ln(x² + 3) + 5, where k < 0. Do the following.(a) Find the critical numbers of f, and make a sign chart for f'(x).(b) Find the open interval(s) on which f is increasing and the open interval(s) onwhich f is decreasing. Justify each answer.(c) Find the value of x where f has a local extreme value, and classify this extremaas a local maximum or minimum. Justify your answer.

Given information: fx=klnx2+3 + 5, where k<0 ------------------ Aim: ------------------------------------------ Basic information: The critical numbers for 'f(x)' are those 'x' values at which either f'(x) = 0 or not defined. Solution: fx=klnx2+3 + 5, where k<0 f'x=kx2+32x use: lngx'=g'xgxaxn'=anxn-1 1--------------------------f'x=0kx2+32x=0 use 1⇒ 2x=0 as k<0 and x2+3>0 for ∀x⇒ x=0 (a critical point) (2)-------------------------- Sign chart for f'(x) = 2xkx2+3, where kx2+3<0For x<0, f'x > 0 as 2x<0 For x>0, f'x < 0 as 2x>0 Basic information: For the given interval , if f'(x)>0, then f(x) is increasing. For the given interval , if f'(x)<0, then f(x) is decreasing. Solution: Interval f'(x) f(x) x<0 or x∈-∞,0 +ve increasing x>0 or x∈0,∞ -ve decreasing Basic information: At local extrema of f(x), f'(x) is equal to zero. Solution: Since f'(x) = 0 at x=0, hence local extrema occurs at x=0. The f(x) is increasing for x<0 and decreasing for x>0. Basically, f(x) goes from increasing to decreasing across x=0 and local extrema at x=0. Hence, f(x) has local maxima at x=0. Let, k = -2.