Question
In Exercises $17-36,$ locate the absolute extrema of the function on the closed interval.$$h(x)=-x^{2}+3 x-5,[-2,1]$$
Step 1
The critical numbers are the solutions of the equation obtained by setting the derivative of the function equal to zero. The derivative of the function $h(x)=-x^{2}+3x-5$ is $h'(x)=-2x+3$. Setting this equal to zero gives $-2x+3=0$, which simplifies to Show more…
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