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Using calculus, find the intervals of increasing, decreasing, concavity, and state the local...

Question:

Using calculus, find the intervals of increasing, decreasing, concavity, and state the local maximum and minimum values as well as inflection points. {eq}f(x)= x^4 + 4x^3 -10{/eq}

Differential Equations

{eq}\text{Critical numbers can be calculated by equating the derivative of the equation to zero.}\\ \text{To list the x-coordinates of all maxima, We have to find }f''(x).\text{ If its value comes out to be positive at critical points.}\\ \text{To list the x-coordinates of all minima, We have to find }f''(x).\text{ If its value comes out to be negative at critical points.}\\ \text{Inflection points are those where }''(x)=0.\\ {/eq}

Answer and Explanation:

{eq}\text{First of all, we will find the local maxima and minima, for that, we have to find critical points }\\ \text{Critical numbers can be calculated by equating the derivative of the equation to zero.}\\ {/eq}

$$f(x)= x^4 + 4x^3 -10\\ f'(x)=4x^3+12x^2\\ \Rightarrow 4x^3+12x^2=0\\ 4x^2(x+3)=0\\ x=0,-3\\ \Rightarrow \text{So, there are two critical points i.e. 0,-3}\\ \text{Now, differentiate again to find maxima and minima.}\\ f''(x)=12x^2+24x\\ f''(0)=0\\ f''(-3)=36\\ f''(x) \text{ is positive which means that local minima is at x=-3 }\\ $$

{eq}\text{For intervals of increasing and decreasing} {/eq}

$$\text{ In the interval }(-3,0)U(0,\infty),\;f'(x)\text{is positive}\\ \therefore \text{the function is increasing in the interval }(-3,0)U(0,\infty)\\ \text{ In the interval }(-\infty,-3),\;f'(x)\text{is negative}\\ \therefore \text{the function is decreasing in the interval }(-\infty,-3)\\ $$

{eq}\text{For inflection points :}f''(x)=0\\ {/eq}

$$12x^2+24x=0\\ 12x(x+2)=0\\ x=0,x=-2\\ \text{There are two inflection points 0,-2 }\\ $$

{eq}\text{For concavity :} {/eq}

$$f''(x)=0 \text{ to find the intervals of concavity}\\ 12x^2+24x=0\\ x=0,x=-2\\ f''x\text{ is positive for}(-\infty,-2)U(0,\infty)\\ \therefore\text{f(x) is concave up for}(-\infty,-2)U(0,\infty)\\ f''x\text{ is negative for}(-2,0)\\ \therefore\text{f(x) is concave down for}(-2,0)\\ $$


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Understanding Concavity and Inflection Points with Differentiation

from Math 104: Calculus

Chapter 10 / Lesson 6
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