Calculus I - Critical Points (Practice Problems)

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Section 4.2 : Critical Points

Determine the critical points of each of the following functions.

$f\left( x \right) = 8{x^3} + 81{x^2} - 42x - 8$ Solution
$R\left( t \right) = 1 + 80{t^3} + 5{t^4} - 2{t^5}$ Solution
$g\left( w \right) = 2{w^3} - 7{w^2} - 3w - 2$ Solution
$g\left( x \right) = {x^6} - 2{x^5} + 8{x^4}$ Solution
$h\left( z \right) = 4{z^3} - 3{z^2} + 9z + 12$ Solution
$Q\left( x \right) = {\left( {2 - 8x} \right)^4}{\left( {{x^2} - 9} \right)^3}$ Solution
$\displaystyle f\left( z \right) = \frac{{z + 4}}{{2{z^2} + z + 8}}$ Solution
$\displaystyle R\left( x \right) = \frac{{1 - x}}{{{x^2} + 2x - 15}}$ Solution
$r\left( y \right) = \sqrt[5]{{{y^2} - 6y}}$ Solution
$h\left( t \right) = 15 - \left( {3 - t} \right){\left[ {{t^2} - 8t + 7} \right]^{\frac{1}{3}}}$ Solution
$s\left( z \right) = 4\cos \left( z \right) - z$ Solution
$\displaystyle f\left( y \right) = \sin \left( \frac{y}{3} \right) + \frac{2y}{9}$ Solution
$V\left( t \right) = {\sin ^2}\left( {3t} \right) + 1$ Solution
$f\left( x \right) = 5x\,{{\bf{e}}^{9 - 2x}}$ Solution
$g\left( w \right) = {{\bf{e}}^{{w^{\,3}} - 2{w^{\,2}} - 7w}}$ Solution
$R\left( x \right) = \ln \left( {{x^2} + 4x + 14} \right)$ Solution
$A\left( t \right) = 3t - 7\ln \left( {8t + 2} \right)$ Solution