Suppose that y(x) denotes a solution of the initial-value problem y'= x^2 + y^2, y(1) =-1 and...

Question:

Suppose that {eq}y(x) {/eq} denotes a solution of the initial-value problem {eq}y'= x^2 + y^2, \ y(1) =-1 {/eq} and that {eq}y(x) {/eq} possesses at least a second derivative at {eq}x = 1 {/eq}. In some neighborhood of {eq}x = 1 {/eq}, use the DE to determine whether {eq}y(x) {/eq} is increasing or decreasing, and whether the graph {eq}y(x) {/eq} is concave up or concave down.

Concavity of a Function:

We are given a real-value function of one variable {eq}y=f(x). {/eq}

The function is concave up in the intervals where its second derivative is positive,

and concave down where the second derivative is negative.

Answer and Explanation:

We know that {eq}y(x) {/eq} is a solution of the initial-value problem

{eq}y'= x^2 + y^2, \ y(1) =-1 {/eq}

and that {eq}y(x) {/eq} admits at least a second derivative at {eq}x = 1 {/eq}.

The second derivative of the function is found by differentiating the former equation by using implicit differentiation

{eq}y''= 2x + 2yy'= 2x + 2y(x^2+y^2) = 2x + 2yx^2 + 2y^3. {/eq}

In the neighborhood of {eq}x = 1, {/eq} the function is increasing if the derivative is positive or decreasing if derivative is negative. Upon using the differential equation

we find that

{eq}y'(1) = 1 + (-1)^2 = 2 > 0 {/eq}

i.e. the first derivative is positive and so the function is increasing.

Moreover, the second derivative is

{eq}y''(1)= 2 + 2(y(1))1^2 + 2(y(1))^3 \\ = 2 + 2(-1) -2 \\ = -2 < 0 {/eq}

and so the function is concave down in the neighborhood of x=1.


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Concavity and Inflection Points on Graphs

from Math 104: Calculus

Chapter 9 / Lesson 5
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