Find all critical points of the function f(x) = 36x^2 \sqrt{64 - x^2}.

Question:

Find all critical points of the function {eq}f(x) = 36x^2 \sqrt{64 - x^2}. {/eq}

Critical Point:

The critical point is term deals with the derivative of a function.

Definition:-

Let f(x) be any differentiable function then the critical point are nothing but the points at which either derivative is zero or not exist at that point.

i.e., the point which satisfies the equation {eq}\displaystyle {f^{'}}(x) = 0. {/eq}

{eq}\displaystyle \textbf{Useful Terms:-} {/eq}

{eq}\displaystyle \eqalign{ & 1)\,\,{\left( {af} \right)^{'}} = a{f^{'}} \cr & 2)\,\,{\left( {f.g} \right)^{'}} = f.{g^{'}} + {f^{'}}g \cr & 3)\,\,{\left( {f \pm g} \right)^{'}} = {f^{'}} \pm {g^{'}} \cr & 4)\,\,\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} \cr} {/eq}

Answer and Explanation:

We have,

{eq}\displaystyle f(x) = 36x^2 \sqrt{64 - x^2}. {/eq}

Now let us differetiate the given function with repsect to x we get,

{eq}\displaystyle \eqalign{ {f^{'}}(x) &= \frac{d}{{dx}}\left( {36{x^2}\sqrt {64 - {x^2}} } \right) \cr & = 36\frac{d}{{dx}}\left( {{x^2}\sqrt {64 - {x^2}} } \right)&\left( {\because {\text{Take}}\:{\text{the}}\:{\text{constant}}\:{\text{out}}} \right) \cr & = 36\left( {\frac{d}{{dx}}\left( {{x^2}} \right)\sqrt {64 - {x^2}} + \frac{d}{{dx}}\left( {\sqrt {64 - {x^2}} } \right){x^2}} \right)&\left( {\because {\text{Apply}}\:{\text{the}}\:{\text{Product}}\:{\text{Rule}}} \right) \cr & = 36\left( {2x\sqrt {64 - {x^2}} + \left( { - \frac{x}{{\sqrt {64 - {x^2}} }}} \right){x^2}} \right) \cr & = \frac{{36\left( { - 3{x^3} + 128x} \right)}}{{\sqrt {64 - {x^2}} }} \cr} {/eq}

The critical points are estimated by equating the first derivative to zero,

{eq}\displaystyle \eqalign{ & {f^{'}}(x) = 0 \cr & \Rightarrow \frac{{36\left( { - 3{x^3} + 128x} \right)}}{{\sqrt {64 - {x^2}} }} = 0\left( {\because {f^{'}}(x) = \frac{{36\left( { - 3{x^3} + 128x} \right)}}{{\sqrt {64 - {x^2}} }}} \right) \cr & \Rightarrow \left( { - 3{x^3} + 128x} \right) = 0 \cr & \Rightarrow x\left( { - {x^2} + 128} \right) = 0 \cr & \Rightarrow x = 0,x = \pm \sqrt {128} \cr} {/eq}

Hence {eq}\displaystyle x = 0,x = \pm \sqrt {128} {/eq} are the critical points for the given function.


Learn more about this topic:

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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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