Solution:
Critical points are the x-values at which the derivative is equal to zero. To find these, find f' as follows:
Set it equal to zero and solve for x:
This is the critical point.
Wednesday, February 22, 2012
Critical Point Problem(Online calculus course test homework solutions)
Question: If a=71 and b=56, then what, to 2 decimal places, is the value of the x-coordinate of the critical point of f? Where the function is:
=\frac{a}{x}+\frac{x^2}{b} "f(x)=\frac{a}{x}+\frac{x^2}{b}")
Solution:
Critical points are the x-values at which the derivative is equal to zero. To find these, find f' as follows:
=\frac{a}{x}+\frac{x^2}{b}=ax^{-1}+(1/b)x^2\Rightarrow f'(x)=-ax^{-2}+(2/b)x "f(x)=\frac{a}{x}+\frac{x^2}{b}=ax^{-1}+(1/b)x^2\Rightarrow f'(x)=-ax^{-2}+(2/b)x")
Set it equal to zero and solve for x:=-ax^{-2}+(2/b)x=0\Rightarrow x^3=\frac{ab}{2}\Rightarrow x=(\frac{ab}{2})^{\frac{1}{3}}\approx12.57 "f'(x)=-ax^{-2}+(2/b)x=0\Rightarrow x^3=\frac{ab}{2}\Rightarrow x=(\frac{ab}{2})^{\frac{1}{3}}\approx12.57")
This is the critical point.
Solution:
Critical points are the x-values at which the derivative is equal to zero. To find these, find f' as follows:
Set it equal to zero and solve for x:
This is the critical point.
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