Question
In Exercises $17-36,$ locate the absolute extrema of the function on the closed interval.$$g(t)=\frac{t^{2}}{t^{2}+3},[-1,1]$$
Step 1
To do this, we need to find the derivative of the function. The derivative of the function $g(t)$ is given by the quotient rule as follows: $$ g'(t)=\frac{(t^{2}+3)(2t)-(t^{2})(2t)}{(t^{2}+3)^{2}}=\frac{2t^{3}+6t-2t^{3}}{(t^{2}+3)^{2}}=\frac{6t}{(t^{2}+3)^{2}} $$ Show more…
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