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Math  /  Calculus

Questionh(z)=1z+9z2 for z>0h(z)=\frac{1}{z}+9 z^{2} \text { for } z>0
Select the exact global maximum and minimum values of the function. The global maximum of h(z)h(z) on z>0z>0 does not exist, the global minimum is 183+943\sqrt[3]{18}+\sqrt[3]{\frac{9}{4}} The global maximum of h(z)h(z) on z>0z>0 is 19+729\frac{1}{9}+729, the global minimum is 93+943\sqrt[3]{9}+\sqrt[3]{\frac{9}{4}} The global maximum of h(z)h(z) on z>0z>0 is 118+729\frac{1}{18}+729, the global minimum is 183+943\sqrt[3]{18}+\sqrt[3]{\frac{9}{4}} The global maximum of h(z)h(z) on z>0z>0 does not exist, the global minimum is 183+923\sqrt[3]{18}+\sqrt[3]{\frac{9}{2}} The global maximum of h(z)h(z) on z>0z>0 is 118+729\frac{1}{18}+729, the global minimum is 93+943\sqrt[3]{9}+\sqrt[3]{\frac{9}{4}}

Studdy Solution

STEP 1

1. The function h(z)=1z+9z2 h(z) = \frac{1}{z} + 9z^2 is defined for z>0 z > 0 .
2. We need to find the critical points by taking the derivative and setting it to zero.
3. Evaluate the function at critical points and analyze behavior as z→0+ z \to 0^+ and z→∞ z \to \infty .

STEP 2

1. Find the derivative of h(z) h(z) .
2. Determine critical points by setting the derivative to zero.
3. Evaluate the function at the critical points.
4. Analyze the behavior of h(z) h(z) as z→0+ z \to 0^+ and z→∞ z \to \infty .
5. Determine the global maximum and minimum values.

STEP 3

Find the derivative of h(z)=1z+9z2 h(z) = \frac{1}{z} + 9z^2 .
h′(z)=−1z2+18z h'(z) = -\frac{1}{z^2} + 18z

STEP 4

Set the derivative equal to zero to find critical points:
−1z2+18z=0 -\frac{1}{z^2} + 18z = 0
18z=1z2 18z = \frac{1}{z^2}
Multiply both sides by z2 z^2 to clear the fraction:
18z3=1 18z^3 = 1
Solve for z z :
z3=118 z^3 = \frac{1}{18}
z=1183 z = \sqrt[3]{\frac{1}{18}}

STEP 5

Evaluate h(z) h(z) at the critical point z=1183 z = \sqrt[3]{\frac{1}{18}} :
h(1183)=11183+9(1183)2 h\left(\sqrt[3]{\frac{1}{18}}\right) = \frac{1}{\sqrt[3]{\frac{1}{18}}} + 9\left(\sqrt[3]{\frac{1}{18}}\right)^2
=183+9×1183 = \sqrt[3]{18} + 9 \times \frac{1}{\sqrt[3]{18}}
=183+943 = \sqrt[3]{18} + \sqrt[3]{\frac{9}{4}}

STEP 6

Analyze the behavior of h(z) h(z) as z→0+ z \to 0^+ and z→∞ z \to \infty :
- As z→0+ z \to 0^+ , 1z→∞ \frac{1}{z} \to \infty and 9z2→0 9z^2 \to 0 , so h(z)→∞ h(z) \to \infty . - As z→∞ z \to \infty , 1z→0 \frac{1}{z} \to 0 and 9z2→∞ 9z^2 \to \infty , so h(z)→∞ h(z) \to \infty .

STEP 7

Determine the global maximum and minimum values:
- The global maximum does not exist because h(z)→∞ h(z) \to \infty as z→0+ z \to 0^+ and z→∞ z \to \infty . - The global minimum is at the critical point z=1183 z = \sqrt[3]{\frac{1}{18}} with value 183+943 \sqrt[3]{18} + \sqrt[3]{\frac{9}{4}} .
The global maximum of h(z) h(z) on z>0 z > 0 does not exist, the global minimum is 183+943 \sqrt[3]{18} + \sqrt[3]{\frac{9}{4}} .

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