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Question Number 11288 by agni5 last updated on 19/Mar/17

Sketch the graph of the function f is defined by f(x)=x^4 +8x^3 , clearly giving all the properties used in it.

$$\mathrm{Sketch}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\:\mathrm{is}\:\mathrm{defined} \\ $$$$\mathrm{by}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{4}} +\mathrm{8x}^{\mathrm{3}} ,\:\mathrm{clearly}\:\mathrm{giving}\:\mathrm{all}\:\mathrm{the}\: \\ $$$$\mathrm{properties}\:\mathrm{used}\:\mathrm{in}\:\mathrm{it}. \\ $$

Answered by ajfour last updated on 19/Mar/17

Commented by ajfour last updated on 19/Mar/17

f(x) = x^4 +8x^3 = x^3 (x+8) at x=0,−8 f(x)=0 f^′ (x) =4x^3 +24x^2 = 4x^2 (x+6) ⇒ f^′ (x) =0 at x=0,0,−6 f^(′′) (x)=12x(x+4) ⇒ point of inflexion at x=0,−4

$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x}^{\mathrm{4}} +\mathrm{8x}^{\mathrm{3}} =\:\mathrm{x}^{\mathrm{3}} \left(\mathrm{x}+\mathrm{8}\right) \\ $$$$\mathrm{at}\:\mathrm{x}=\mathrm{0},−\mathrm{8}\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\mathrm{f}^{'} \left(\mathrm{x}\right)\:=\mathrm{4x}^{\mathrm{3}} +\mathrm{24x}^{\mathrm{2}} \:=\:\mathrm{4x}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{6}\right) \\ $$$$\Rightarrow\:\mathrm{f}^{'} \left(\mathrm{x}\right)\:=\mathrm{0}\:\:\mathrm{at}\:\mathrm{x}=\mathrm{0},\mathrm{0},−\mathrm{6} \\ $$$$\mathrm{f}^{''} \left(\mathrm{x}\right)=\mathrm{12x}\left(\mathrm{x}+\mathrm{4}\right) \\ $$$$\Rightarrow\:\mathrm{point}\:\mathrm{of}\:\mathrm{inflexion}\:\mathrm{at}\:\mathrm{x}=\mathrm{0},−\mathrm{4} \\ $$

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