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Question Number 78890 by gopikrishnan last updated on 21/Jan/20

$$thelocalmaximumvalueoff\left(x\right)=x^4+32x$$

Commented by mr W last updated on 21/Jan/20

$$thereisnolocalmaximum,only$$$$local(=global)minimumatx=-2.$$

Commented by mathmax by abdo last updated on 22/Jan/20

$$f\left(x\right)=x^4+32\Rightarrow f^{{}^{\prime }}\left(x\right)=4x^3+32=4(x^3+8)=4(x+2)(x^2-2x+4)$$$$f^{{}^{\prime }}\left(x\right)⩾0\iff x⩾-2$$$$x-\mathrm{\infty }-2+\mathrm{\infty }$$$$f^{{}^{\prime }}-+$$$$f+\mathrm{\infty }decrf(-2)incr+\mathrm{\infty }$$$$nomaxforfandminf\left(x\right)=f(-2)={(-2)}^4-64=16-64=-48$$$$$$

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