Use f^prime(x) and f^prime \prime(x) to draw the graph of f(x)=9-x^2. Use f^prime(x) and f^prime \prime(x)


  1. Use \(f^{\prime}(x)\) and \(f^{\prime \prime}(x)\) to draw the graph of \(f(x)=9-x^{2}\).
  2. Use \(f^{\prime}(x)\) and \(f^{\prime \prime}(x)\) to draw the graph of \(f(x)=4 x^{3}-6 x^{2}+5\)
  3. Draw the graph of \(f(x)=e^{-x^{2}}\). Here \(f^{\prime}(x)=-2 x e^{-x^{2}}\) and \(f^{\prime \prime}(x)=2\left(2 x^{2}-1\right) e^{-x^{2}}\).
  4. Draw the graph of \(f(x)=\ln \left(x^{2}+1\right)\). Here \(f^{\prime}(x)=\frac{2 x}{x^{2}+1}\) and \(f^{\prime \prime}(x)=\frac{2-2 x^{2}}{\left(x^{2}+1\right)^{2}}\).
  5. Find the maximum and minimum of \(f(x)=4 x^{3}-6 x^{2}+5\) on the intervals (a) \(-1 \leq x \leq 1\),
    (b) \(1 \leq x \leq 2\), (c) \(0 \leq x \leq 1\), and (d) \(-1 / 2 \leq x \leq 3 / 2\). (See problem #2 above.)
  6. Find the average value of the function \(f(x)=x^{2}+1\) on the interval \(-1 \leq x \leq 5\).
  7. Find the average value of the function \(f(x)=\frac{3 x^{2}}{x^{3}+1}\) on the interval \(0 \leq x \leq 2\).
  8. Find the average value of the function \(f(x)=e^{x}\) on the interval \(0 \leq x \leq \ln 2\).
  9. Find the area bounded by the graph of \(f(x)=9-x^{2}\) and the \(x\) -axis.
  10. The graphs of \(f(x)=x^{2}\) and \(g(x)=x\) intersect at the points \((0,0)\) and \((1,1)\). Find the area between the two graphs.
  11. Approximate \(\sqrt[4]{15.9}\) by using a linear function tangent to the graph of \(f(x)=\sqrt[4]{x}\)
  12. Find the piecewise linear function of the form

\(f(x)=\left\{\begin{array}{lc}

? & -2 \leq x \leq-1 \\

? & -1 \leq x \leq 0 \\

? & 0 \leq x \leq 1 \\

? & 1 \leq x \leq 2 \\

? & 2 \leq x \leq 3

\end{array}\right.\)

which approximates \(g(x)=\frac{100 x}{\left(x^{2}+1\right)^{2}}\). Find the average value of both \(f(x)\) and \(g(x)\) on the interval \(-2 \leq x \leq 3 .\) Here \(\int \frac{100 x}{\left(x^{2}+1\right)^{2}} d x=-\frac{50}{x^{2}+1}\)

Price: $18.82
Solution: The downloadable solution consists of 13 pages, 582 words and 5 charts.
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