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Question Number 95643 by Ar Brandon last updated on 26/May/20

Show that the function f(x)=x^3 is derivable at all points x_0 ∈R and that f′(x_0 )=3x_0 ^2

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} \:\mathrm{is}\:\mathrm{derivable}\:\mathrm{at}\:\mathrm{all} \\ $$$$\mathrm{points}\:\mathrm{x}_{\mathrm{0}} \in\mathbb{R}\:\mathrm{and}\:\mathrm{that}\:\mathrm{f}'\left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{3x}_{\mathrm{0}} ^{\mathrm{2}} \\ $$

Answered by Rio Michael last updated on 26/May/20

lets choose a general point c ∈ R , if f(x) = x^3 is derivable or differentiable at x = x_0 = c then lim_(x→c) ((f(x)−f(c))/(x−c)) must exist. but f(x) = x^3 ⇒ f(c) = c^3 lim_(x→c) ((x^3 −c^3 )/(x−c)) = lim_(x→c) (((x−c)(x^2 + 2cx + c^3 ))/((x−c))) = lim_(x→c) (x^2 + 2cx + c^2 ) = 3c^2

$$\mathrm{lets}\:\mathrm{choose}\:\mathrm{a}\:\mathrm{general}\:\mathrm{point}\:{c}\:\in\:\mathbb{R}\:, \\ $$$$\mathrm{if}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:\mathrm{is}\:\mathrm{derivable}\:\mathrm{or}\:\mathrm{differentiable}\:\mathrm{at}\:{x}\:=\:{x}_{\mathrm{0}} \:=\:{c}\:\mathrm{then} \\ $$$$\underset{{x}\rightarrow{c}} {\mathrm{lim}}\frac{{f}\left({x}\right)−{f}\left({c}\right)}{{x}−{c}}\:\mathrm{must}\:\mathrm{exist}. \\ $$$$\mathrm{but}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:\Rightarrow\:{f}\left({c}\right)\:=\:{c}^{\mathrm{3}} \\ $$$$\:\underset{{x}\rightarrow{c}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}} −{c}^{\mathrm{3}} }{{x}−{c}}\:=\:\underset{{x}\rightarrow{c}} {\mathrm{lim}}\:\frac{\left({x}−{c}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{2}{cx}\:+\:{c}^{\mathrm{3}} \right)}{\left({x}−{c}\right)}\:=\:\underset{{x}\rightarrow{c}} {\mathrm{lim}}\:\left({x}^{\mathrm{2}} \:+\:\mathrm{2}{cx}\:+\:{c}^{\mathrm{2}} \right)\:=\:\mathrm{3}{c}^{\mathrm{2}} \\ $$$$\:\:\: \\ $$

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