Let g: (a, b) \rightarrow R and let x_0 \epsilon (a, b). x_0 is a critical number? g is...

Question:

Let {eq}g{/eq}: {eq}(a, b) \rightarrow R{/eq} and let {eq}x_0 \epsilon (a, b){/eq}.

{eq}x_0{/eq} is a critical number?

{eq}g{/eq} is decreasing or increasing on {eq}(a, b){/eq}

Critical Point:

{eq}\\ {/eq}

For a differentiable function, say {eq}f {/eq} with variable {eq}x {/eq}, a critical point is such a point defined on the function where the first derivative of the function is zero i.e. simply we have to differentiate the function and equate to zero to obtain all such critical points {eq}x_{0}, x_{1},......x_{n} {/eq} .

Answer and Explanation:

{eq}\\ {/eq}

For a function {eq}g{/eq}: {eq}(a, b) \rightarrow R{/eq}, {eq}x_{0} {/eq} i.e the critical point is such a point where the tangent line is horizontal or we can say, the derivative is zero at that point.

We have to differentiate {eq}g {/eq} to obtain {eq}g' {/eq} and equate {eq}g'(x)=0 {/eq} to find the critical points.

Let {eq}h_{0} {/eq} be the point on the leftmost of the critical point {eq}x_{1} {/eq} and {eq}h_{n} {/eq} be the point to the rightmost of the critical point {eq}x_{n} {/eq}.

So, on the outside interval {eq}(-\infty, x_{0}) {/eq}, the function is increasing on {eq}(a,b) {/eq} if {eq}g'(h_{0})>0 {/eq} and decreasing on {eq}(a,b) {/eq} if {eq}g'(h_{0})<0 {/eq}. In the same way, {eq}(\infty, x_{0}) {/eq}, the function is increasing on {eq}(a,b) {/eq} if {eq}g'(h_{n})>0 {/eq} and decreasing on {eq}(a,b) {/eq} if {eq}g'(h_{n})<0 {/eq}.


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
196K

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