Answer to Question #123126 in Real Analysis for Wachira Ann Wangari

Question #123126
Let f(x)=|x|^3. Show that f'''(0) does not exist
1
Expert's answer
2020-06-22T17:18:38-0400

"f(x)=|x|^3=\\left\\{\\begin{array}{ll}x^3 & \\text { if } x \\geq 0 \\\\ -x^3 & \\text { if } x<0\\end{array}\\right.\\\\[1 em]\nSo, \\lim _{x \\rightarrow 0^{+}}|x|^3=\\lim _{x \\rightarrow 0^{+}} x^3=0\n~~\\text{and} ~~\\lim _{x \\rightarrow 0^{-}}|x|^3=\\lim _{x \\rightarrow 0^{-}}(-x)^3=0\\\\[1 em]\n\n\n\\therefore ~f(x)=|x|^3~~\\text{is continuous at x=0}\\\\[1 em]\n\\text{To show that} f(x)=|x|^3 ~\\text{is not differentiable} \\\\[1 em]\n\n\n\nf^{\\prime}(0)=\\lim _{h \\rightarrow 0} \\frac{f(0+h)-f(0)}{h}\\\\[1 em]\n\n\n\\begin{array}{l}\n\\lim _{h \\rightarrow 0} \\frac{|0+h|^3-|0|^3}{h}=\\lim _{h \\rightarrow 0} \\frac{|h|^3}{h} \n=\\left\\{\\begin{array}{ll}\n0 & \\text { if } h>0 \\\\\n0 & \\text { if } h<0\n\\end{array}\\right.\n\\end{array}\\\\[1 em]\n\n\n\n\n\\therefore ~f^{\\prime}(x) \\text{ exist at }x=0\\\\[1 em]\nf^{{\\prime }}(x)=\\left\\{\\begin{array}{ll}3x^2 & \\text { if } x \\geq 0 \\\\ -3x^2 & \\text { if } x<0\\end{array}\\right.\\\\[1 em]\n\\text{In the same way, we find that the second derivative is exist at}~x=0\\\\[1 em]\nf^{{\\prime }{\\prime }}(x)=\\left\\{\\begin{array}{ll}6x & \\text { if } x \\geq 0 \\\\ -6x & \\text { if } x<0\\end{array}\\right.\\\\[1 em]\n\\text{Now we are checking the third derivative at}~x=0\\\\[1 em]\nf^{{\\prime }{\\prime }{\\prime }}(0)^{+}=\\lim _{x \\rightarrow 0^{+}}f^{{\\prime }{\\prime }}(x)= \\frac{6(0+h)-f(0)}{h}=6 \\\\[1 em]\n\nf^{{\\prime }{\\prime }{\\prime }}(0)^{-}=\\lim _{x \\rightarrow 0^{-}}f^{{\\prime }{\\prime }}(x)= \\frac{-6(0+h)-f(0)}{h}=-6 \\\\[1 em]\n\n\n\n\n\n\n\\therefore~f^{{\\prime }{\\prime }{\\prime }}(0) \\text{does not exist}"

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