How do you determine whether the function f(x)=4x2+1 is concave up or concave down and its intervals?

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Juan Goodwin · Accepted Answer

Step 1If a function is differentiable twice, we know that it is concave up if&amp;nbsp;fx)&amp;gt;0&amp;nbsp;and concave down viceversa.So let us calculate:f(x)=4x2+1=4(x2+1)−2f′(x)=−16x(x2+1)−3fx)=96x2(x2+1)−4−16(x2+1)−3=96x2−16(x2+1)(x2+1)4=96x2−16x2−16(x2+1)4=165x2−1(x2+1)4Step 2Evidently the denominator of f&quot;(x) is always positive, so the sign of f&quot;(x) is the sign of the numerator and we see that:fx)&amp;lt;0&amp;nbsp;for&amp;nbsp;|x|&amp;lt;15So f(x) is concave up in the intervals&amp;nbsp;(−∞,−15&amp;nbsp;and&amp;nbsp;(15,+∞)&amp;nbsp;&amp;nbsp;and concave down for x in the interval&amp;nbsp;(-15,15).

How do you determine whether the function

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