How do you determine where the graph of the given function is increasing, decreasing, concave up, and concave down for h(x)=x2x2+1?

Question

xxkrnjangxxed9q · Accepted Answer

Step 1First we need to evaluate the domain and the first and second derivatives:Dh=′Rh′(x)=2x(x2+1)2hx)=−2(3x2−1)(x2+1)3=−6(x+33)(x−33)(x2+1)3Step 2Now, h is increasing when h′&amp;gt;0 and decreasing when h′&amp;lt;0. Notice that ∀x∈R x2+1&amp;gt;0 so as long and we&#039;re interested only in the sing of the derivative we can ommit the denominator.2x&amp;gt;0⇔x&amp;gt;0 - function h increases when x∈(0;+∞)2x&amp;lt;0⇔x&amp;lt;0 - function h decreases when x∈(−∞;0)Function h is concave up when h0 and concave down when h0.concave up: −6(x+33)(x−33)&amp;gt;0⇔x∈(−33;33)concave down: −6(x+33)(x−33)&amp;lt;0⇔x∈(−∞;−33)∪(33;+∞)

How do you determine where the graph of

Answered question

Answer & Explanation

New Questions in Calculus 1