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Question Number 78890 by gopikrishnan last updated on 21/Jan/20
the local maximum value of f(x)=x^4 +32x
thelocalmaximumvalueoff(x)=x4+32x
Commented by mr W last updated on 21/Jan/20
there is no local maximum, only local (=global) minimum at x=−2.
thereisnolocalmaximum,onlylocal(=global)minimumatx=−2.
Commented by mathmax by abdo last updated on 22/Jan/20
f(x)=x^4 +32 ⇒f^′ (x)=4x^3 +32 =4(x^3 +8) =4(x+2)(x^2 −2x+4) f^′ (x)≥0 ⇔x≥−2 x −∞ −2 +∞ f^′ − + f +∞ decr f(−2) incr +∞ no max forf and minf(x)=f(−2) =(−2)^4 −64 =16−64=−48
f(x)=x4+32⇒f′(x)=4x3+32=4(x3+8)=4(x+2)(x2−2x+4)f′(x)⩾0⇔x⩾−2x−∞−2+∞f′−+f+∞decrf(−2)incr+∞nomaxforfandminf(x)=f(−2)=(−2)4−64=16−64=−48

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