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[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 2, Section 2.1

See attachment.

Enjoy.

You may do both or select your favorite to answer. Isn't math great?

[Math Tutor]

98. [tex]A(x)=4x\sqrt{1-x^2}[/tex]
(a) [tex]A\left(\tfrac{1}{3}\right)=\tfrac{4}{3}\sqrt{\tfrac{8}{9}}=\tfrac{8\sqrt{2}}{9}\approx 1.26\ \text{ft}^2[/tex]
(b) [tex]A\left(\tfrac{1}{2}\right)=2\sqrt{\tfrac{3}{4}}=\sqrt{3}\approx 1.73\ \text{ft}^2[/tex]
(c) [tex]A\left(\tfrac{2}{3}\right)=\tfrac{8}{3}\sqrt{\tfrac{5}{9}}=\tfrac{8\sqrt{5}}{9}\approx 1.99\ \text{ft}^2[/tex]

[nycmath]

98. [tex]A(x)=4x\sqrt{1-x^2}[/tex]
(a) [tex]A\left(\tfrac{1}{3}\right)=\tfrac{4}{3}\sqrt{\tfrac{8}{9}}=\tfrac{8\sqrt{2}}{9}\approx 1.26\ \text{ft}^2[/tex]
(b) [tex]A\left(\tfrac{1}{2}\right)=2\sqrt{\tfrac{3}{4}}=\sqrt{3}\approx 1.73\ \text{ft}^2[/tex]
(c) [tex]A\left(\tfrac{2}{3}\right)=\tfrac{8}{3}\sqrt{\tfrac{5}{9}}=\tfrac{8\sqrt{5}}{9}\approx 1.99\ \text{ft}^2[/tex]

Ok. This is just plug and chug.

[Eigenvalue]

99. R(x)=([tex]\frac{L}{P}[/tex])(x)

[nycmath]

99. R(x)=([tex]\frac{L}{P}[/tex])(x)

Correct.

We can also write it this way:

R(x) = L(x)/P(x)