For $s(x)={\left(x+9\right)}^4$, what does the Second Derivative Test say about the critical point at $x=-9$?

A student writes: $f^{\prime }\left(-8\right)=0$ and $f^{\prime \prime }\left(-8\right)=0$, so $f$ has a relative minimum at $x=-8$. Which best evaluates the argument?

For $f(x)=x\ln x$ with $x>0$, classify the critical point at $x=\frac{1}{e}$ using the Second Derivative Test.

For $s(x)=2x^3-6x^2-18x$, use the Second Derivative Test to classify the critical point at $x=3$.

For $g(x)=2x^3-9x^2-24x$, use the Second Derivative Test to classify the critical point at $x=4$.

For $f(x)={\left(x+6\right)}^3$, what does the Second Derivative Test say about the critical point at $x=-6$?

For $f(x)=\sin x$ with $x$ in radians, classify the critical point at $x=\frac{\pi }{2}$ using the Second Derivative Test.

For $w(x)=2x^3+3x^2-36x$, use the Second Derivative Test to classify the critical point at $x=-3$.

For $f(x)=\cos x$ with $x$ in radians, classify the critical point at $x=\pi$ using the Second Derivative Test.

A function has $f^{\prime }(7)=0$ and $f^{\prime \prime }(7)=0$. What does the Second Derivative Test give?

For $v(x)=2x^3-12x^2$, use the Second Derivative Test to classify the critical point at $x=4$.

For $f(x)=x{e}^x$, classify the critical point at $x=-1$ using the Second Derivative Test.

A student writes: $r^{\prime }\left(2\right)=0$ and $r^{\prime \prime }\left(2\right)=5$, so $r$ has a relative maximum at $x=2$. Which best evaluates the argument?

For $q(x)={\left(x-4\right)}^3$, what does the Second Derivative Test say about the critical point at $x=4$?

A function has $f^{\prime }(4)=3$ and $f^{\prime \prime }(4)=-2$. What does the Second Derivative Test say about $x=4$?

For $f(x)=2x^3-12x^2$, use the Second Derivative Test to classify the critical point at $x=0$.

For $f(x)=e^x-x$, classify the critical point at $x=0$ using the Second Derivative Test.

A function has a critical point at $x=3$ where its first derivative is UNDEFINED. Can the Second Derivative Test classify it?

For $f(x)=2x^3+9x^2+12x$, use the Second Derivative Test to classify the critical point at $x=-2$.

For $p(x)=2x^3+18x^2+48x$, use the Second Derivative Test to classify the critical point at $x=-4$.

For $u(x)=2x^3+15x^2+36x$, use the Second Derivative Test to classify the critical point at $x=-2$.

For $p(x)={\left(x-2\right)}^4$, what does the Second Derivative Test say about the critical point at $x=2$?

For $f(x)={\left(x+3\right)}^3$, what does the Second Derivative Test say about the critical point at $x=-3$?

A student writes: $u^{\prime }\left(4\right)=0$ and $u^{\prime \prime }\left(4\right)=0$, so $u$ has a relative minimum at $x=4$. Which best evaluates the argument?