# find the local maximum and minimum values of f using both the first and second derivative tests. f(x)=x^5-5x+6

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The maximum and minimum are the solutions of f'(x) = 0, for which f"(x) < 0 or f"(x) > 0.

So we first find the derivative f'(x) and set f'(x) = 0 and solve for x.

f'(x) = (x^5-5x+6)'.

f'(x) = 5x^4-5.

f'(x) = 0 => 5x^4-5 = 0

=> (5(x^4-1) = 0.

=> (x^4-1) = 0.

=> (x^2-1)(x+1) = 0.

=>(x-1)(x+1)(x^2+1) = 0.

=> x-1 = 0 or x+1 = 0 or x^2+1.

=> x= 1 or x = -1 are real solutions.

f"(x) = f''(x) = (5x^4-5)' = 5*4*x^3 = 20x^3.

f"(-1) = 20(-1)^3 <0. So at x = -1, f(x) = (x^5-5x+6) is maximum.

f"(+1) = 20*(1)^3 >0. So for x = 1, f(x) = (x^5-5x+6) is minimum.

So when x= 1, f(-1) = (-1)^5-5(-1)+6 = 10 is maximum.

When x= 1, f(1) = (1)^5-5+6 = 2 is minimum.

never mind i got it, its 2 and 10