Answer to Question #192494 in Real Analysis for Nikhil

Question

Check whether the function f given by:

f(x)= (x-4)^2(x+1)^4 has local maxima and loca

minima

Answer 1

Solution

f’(x) = 2(x-4) (x+1)⁴ +4(x-4)²(x+1)³ = 2(x-4) (x+1)³ [x+1+2(x-4)] = 2(x-4) (x+1)³ (3x-7)

So solutions of the equation f’(x) = 0 are x₁ = -1, x₂ = 7/3, x₃ = 4. These points are points of extremum.

f’’(x) = 2 (x+1)³ (3x-7)+6(x-4) (x+1)² (3x-7)+6(x-4) (x+1)³  = 2 (x+1)² [(x+1)(3x-7)+3(x-4)(3x-7)+ 3(x-4) (x+1)] =  2 (x+1)² [3x²-4x-7+3(x-4)(4x-6)] = 2 (x+1)² [3x²-4x-7+12x² -66x+72] = 2 (x+1)² (15x²-70x+65)

f’’(x₁) = 0, f’’(x₂) = -370.37 <0, f’’(x₃) = 1250>0.

Therefore x₃=4 is a point of local minima, x₂=7/3 is a point of local maxima

For point x₁=-1 f(x₁)=0 and for any ε>0 (x₁± ε +1)⁴ >0 and f(x₁± ε)>0. So x₁=-1 is a point of local minima too.    

Answer

x₁=-1 and x₃=4 are points of local minima, x₂=7/3 is a point of local maxima