Question to test your Finding Global Extrema knowledge

Question

Find the critical points, the local maximum value, and the local minimum value of the function $h(x) = \frac{7 - x}{x^2 + 3x - 10}$ on the interval $\left\lbrack-10,10\right\rbrack$ . Also, check for the absolute maximum and the absolute minimum, if they exist.

Accepted Answer

Critical point: $h(x) = \frac{7 - x}{x^2 + 3x - 10}$ ,

Local maximum: $h(7 - 2\sqrt{15}) = -\frac{4\sqrt{15} + 17}{49}$ ,

No local minimum,

No absolute maximum or minimum exists on $[-10, 10]$ .

Answer

Critical point: $(7 - 2 \sqrt{15}, \frac{4 \sqrt{15} + 17}{49})$ ,

Local minimum: $h(7 - 2\sqrt{15}) = -\frac{4\sqrt{15} + 17}{49}$ ,

No local maximum,

No absolute maximum or minimum exists on $[-10, 10]$ .

Answer

Critical point: $(7 - 2 \sqrt{15}, \frac{4 \sqrt{15} + 17}{49})$ ,

Local maximum: $h(7 - 2\sqrt{15}) = -\frac{4\sqrt{15} + 17}{49}$ ,

No local minimum,

Absolute maximum: $h\left(10\right)=-\frac{1}{40}$ ,

Absolute minimum: $h\left(-10\right)=\frac{17}{60}$ .

Answer

Critical point: $h(x) = \frac{7 - x}{x^2 + 3x - 10}$ ,

Local maximum: $h(7 - 2\sqrt{15}) = -\frac{4\sqrt{15} + 17}{49}$ ,

No local minimum,

Absolute maximum: $h (- 10) = \frac{17}{60}$ ,

Absolute minimum: $h (10) = - \frac{1}{40}$ .

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