# Chapter 3 - Applications of Differentiation - 3.1 Exercises: 25

Over the specified interval, the function has an absolute maximum equal to $\dfrac{1}{4}$ and an absolute minimum equal to $0$.

#### Work Step by Step

Using the quotient rule: $g'(t)=(\frac{u(t)}{v(t)})'=\frac{u'(t)v(t)-v'(t)u(t)}{(v(t))^2}$. $u(t)=t^2 ;u'(t)=2t$. $v(t)=t^2+3 ;v'(t)=2t$ $g'(t)=\dfrac{(2t)(t^2+3)-(2t)(t^2)}{(t^2+3)^2}=\dfrac{6t}{(t^2+3)^2}.$ $g'(t)=0\to 6t=0\to t=0$ $g'(t)$ is defined for all values of $t.$ Along with the interval endpoints, $t=0$ is a possible value at which the function might attain an absolute extremum. $g(-1)=\dfrac{1}{4}.$ $g(0)=0.$ $g(1)=\dfrac{1}{4}$ Over the specified interval, the function has an absolute maximum equal to $\dfrac{1}{4}$ and an absolute minimum equal to $0$.

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