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Practice Problem Solutions for 2/25

  1. Consider the function \(f(x)=\sqrt[3]{x}\)
    1. Show that \(f\) is continuous at the point \(x=0\).

      2. \(f(0)=0\), and by the root rule, \(\lim_{x\rightarrow 0}\sqrt[3]{x}=0\). Since these agree, \(f\) is continuous at 0 by the continuity test.
    2. Use the limit definition of a derivative to determine whether or not \(f(x)\) is differentiable at \(x=0\).
      3. \[ \lim_{h\rightarrow 0} \frac{\sqrt[3]{0+h}-\sqrt[3]{0}}{h}=\lim_{h\rightarrow 0}\frac{h^{1/3}}{h}=\lim_{h\rightarrow 0}\frac{1}{h^{2/3}} \] This limit does not exist as it tends to infinity. Therfore, \(f(x)=\sqrt[3]{3}\) is not differentiable at the point \(x=0\). (FYI it is differentiable everywhere else.)
      This example shows that there are functions which are continuous but not differentiable. However, we know what if a function is differentiable, it must be continuous.
    3. What does the graph of the function look like in a small interval around \(x=0\)?
      4. The graph is getting steeper and steeper so that the tangent line at the point 0 is vertical.

  2. Find the derivative of \(f(x)= 2x^{-2/5}+\frac{x^5+3x^2+4}{x^{11}+9x^7}+5\)

    3. Assuming each term is differentialbe, we can first apply the sum rule: \[ \frac{d}{dx}f(x)=\frac{d}{dx}2x^{-2/5}+\frac{d}{dx}\frac{x^5+3x^2+4}{x^{11}+9x^7}+\frac{d}{dx}5. \] Using both the power rule and constant multiple rule, the derivative of the first term is \(\frac{-4}{5x^{7/5}}\).
    We can apply the quotient rule to find the derivative of the second term \[ \frac{d}{dx}\frac{x^5+3x^2+4}{x^{11}+9x^7}=\frac{(5x^4+6x)(x^{11}+9x^7)-(x^5+3x^2+4)(11x^{10}+63x^6)}{(x^{11}+9x^7)^2} \] Since \(\frac{d}{dx}5=0\) we have \[ \frac{d}{dx}f(x)=\frac{-4}{5x^{7/5}}-\frac{6x^{15}+33x^{12}+18x^{11}+44x^{10}+189x^8-252x^6}{(x^{11}+9x^7)^2} \]