# Consider the following function: f(x) = 6x + 3 \: sin(x). Find all critical points in the graph...

## Question:

Consider the following function: {eq}f(x) = 6x + 3 \: sin(x) {/eq}. Find all critical points in the graph of the function.

## Critical Points:

The critical point is also referred to as the stationary point, and the tangent line to the graph or curve is horizontal at the critical point.

The tangent's slope is determined using the function's first-order derivative. So, we find the tangent's slope and equate it to zero to identify the critical points.

We are given the following data:

• The given function is: {eq}f\left( x \right) = 6x + 3\sin x {/eq}

Differentiate the given function with respect to x.

{eq}\begin{align*} f'\left( x \right) &= \dfrac{d}{{dx}}\left( {6x + 3\sin x} \right)\\ &= 6\dfrac{d}{{dx}}\left( x \right) + 3\dfrac{d}{{dx}}\left( {\sin x} \right)&\left[ {{\rm{Use}}\;{\rm{Sum}}\;{\rm{and}}\;{\rm{Constant}}\;{\rm{Rule}}} \right]\\ &= 6\left( 1 \right) + 3\cos x&\left[ {{\rm{Use}}\;{\rm{Power}}\;{\rm{Rule}}} \right]\\ &= 6 + 3\cos x \end{align*} {/eq}

Equate the first-order derivative to zero to find the critical points.

{eq}\begin{align*} f'\left( x \right) &= 0\\ 6 + 3\cos x &= 0\\ 3\left( {2 + \cos x} \right) &= 0\\ 2 + \cos x &= 0\\ \cos x &= - 2 \end{align*} {/eq}

From the above calculation, it is clear that the function has no critical point because there is no solution to the above equation because {eq}- 1 \le \cos x \le 1. {/eq}

 Thus, there is no critical point. Finding Critical Points in Calculus: Function & Graph

from

Chapter 8 / Lesson 9
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This lesson explores what critical points are in calculus. It gives a step-by-step explanation of how to find the critical points of a function, and it explains the significance of these points.