How do you find local maximum value of f using the first and second derivative tests: $f(x) = x^5 - 5x + 5$ ?

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Answer 1 · 2016-09-03T15:31:26

$f(-1)=-1+5+5=9" is local maximum"$ .

It is known from Calculus that a fun. $f : RRrarrRR$ has a local

maxima at $x=c", then,":(1): f'(c)=0, &, :(2): f''(c)<0$ .

$f(x)=x^5-5x+5 rArr f'(x)=5x^4-5, and, f''(x)=20x^3$ .

$"Now, "f'(x)=0rArr5x^4-5=0rArrx=+-1$

Since, $f''(-1)=-20<0,$ , we find that,

$f(-1)=-1+5+5=9" is local maximum"$ .

How do you find local maximum value of f using the first and second derivative tests: f(x) = x^5 - 5x + 5f(x) = x^5 - 5x + 5?