Math / Calculus

QuestionConsider the following function. $h(z)=\frac{1}{z}+5 z^{2} \text { for } z>0$
Select the exact global maximum and minimum values of the function. The global maximum of $h(z)$ on $z>0$ is $\frac{1}{10}+125$ , the global minimum is $\sqrt[3]{10}+\sqrt[3]{\frac{5}{4}}$ The global maximum of $h(z)$ on $z>0$ does not exist, the global minimum is $\sqrt[3]{10}+\sqrt[3]{\frac{5}{2}}$ The global maximum of $h(z)$ on $z>0$ is $\frac{1}{10}+125$ , the global minimum is $\sqrt[3]{5}+\sqrt[3]{\frac{5}{4}}$ The global maximum of $h(z)$ on $z>0$ does not exist, the global minimum is $\sqrt[3]{10}+\sqrt[3]{\frac{5}{4}}$ The global maximum of $h(z)$ on $z>0$ is $\frac{1}{5}+125$ , the global minimum is $\sqrt[3]{5}+\sqrt[3]{\frac{5}{4}}$ eTextbook and Media

Studdy Solution

STEP 1

What is this asking? Find the biggest and smallest values of $h(z) = \frac{1}{z} + 5z^2$ for positive $z$ . Watch out! Don't forget to check what happens when $z$ gets really close to zero and really, really big!

STEP 2

1. Explore the function's behavior
2. Find the minimum
3. Find the maximum

STEP 3

Let's see what happens when $z$ gets super close to zero.
As $z$ approaches zero, $\frac{1}{z}$ becomes gigantic, so $h(z)$ explodes to infinity!

STEP 4

Now, what happens when $z$ becomes huge?
Well, $5z^2$ also becomes massive, so $h(z)$ goes to infinity again!

STEP 5

To find the minimum, we need to find the derivative of $h(z)$ and set it to zero!
The derivative tells us how $h(z)$ is changing.

STEP 6

$\frac{d}{dz} h(z) = \frac{d}{dz} \left( \frac{1}{z} + 5z^2 \right) = -\frac{1}{z^2} + 10z$

STEP 7

Now, set the derivative equal to zero and solve for $z$ : $-\frac{1}{z^2} + 10z = 0$ Multiply both sides by $z^2$ (since $z > 0$ , we're not multiplying by zero, so it's all good!): $-1 + 10z^3 = 0$ $10z^3 = 1$ $z^3 = \frac{1}{10}$ $z = \sqrt[3]{\frac{1}{10}}$

STEP 8

Plug this critical value back into $h(z)$ to find the minimum value: \begin{align} h\left(\sqrt[3]{\frac{1}{10}}\right) &= \frac{1}{\sqrt[3]{\frac{1}{10}}} + 5\left(\sqrt[3]{\frac{1}{10}}\right)^2 \\ &= \sqrt[3]{10} + 5\left(\frac{1}{10}\right)^{\frac{2}{3}} \\ &= \sqrt[3]{10} + 5\left(\frac{1}{100}\right)^{\frac{1}{3}} \\ &= \sqrt[3]{10} + 5 \cdot \frac{1}{\sqrt[3]{100}} \\ &= \sqrt[3]{10} + 5 \cdot \frac{1}{\sqrt[3]{4 \cdot 25}} \\ &= \sqrt[3]{10} + 5 \cdot \frac{1}{\sqrt[3]{4} \cdot \sqrt[3]{25}} \cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}} \cdot \frac{\sqrt[3]{5}}{\sqrt[3]{5}} \\ &= \sqrt[3]{10} + \sqrt[3]{\frac{5 \cdot 5 \cdot 5}{4 \cdot 5 \cdot 4}} \\ &= \sqrt[3]{10} + \sqrt[3]{\frac{125}{20 \cdot 4}} = \sqrt[3]{10} + \sqrt[3]{\frac{125}{80}} = \sqrt[3]{10} + \sqrt[3]{\frac{5}{4} \cdot \frac{5}{4} \cdot \frac{5}{2}} = \sqrt[3]{10} + \sqrt[3]{\frac{5}{4} \cdot \frac{25}{8}} \\ &= \sqrt[3]{10} + \sqrt[3]{\frac{5}{4}} \end{align}

STEP 9

We already saw that $h(z)$ goes to infinity as $z$ gets close to zero or becomes very large.
This means there's no global maximum!

STEP 10

The global maximum of $h(z)$ does not exist.
The global minimum is $\sqrt[3]{10} + \sqrt[3]{\frac{5}{4}}$ .

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Studdy Solution

STEP 1

What is this asking? Find the biggest and smallest values of $h(z) = \frac{1}{z} + 5z^2$ for positive $z$ . Watch out! Don't forget to check what happens when $z$ gets really close to zero and really, really big!

STEP 2

1. Explore the function's behavior
2. Find the minimum
3. Find the maximum

STEP 3

Let's see what happens when $z$ gets super close to zero.
As $z$ approaches zero, $\frac{1}{z}$ becomes gigantic, so $h(z)$ explodes to infinity!

STEP 4

Now, what happens when $z$ becomes huge?
Well, $5z^2$ also becomes massive, so $h(z)$ goes to infinity again!

STEP 5

To find the minimum, we need to find the derivative of $h(z)$ and set it to zero!
The derivative tells us how $h(z)$ is changing.

STEP 6

$\frac{d}{dz} h(z) = \frac{d}{dz} \left( \frac{1}{z} + 5z^2 \right) = -\frac{1}{z^2} + 10z$

STEP 7

Now, set the derivative equal to zero and solve for $z$ : $-\frac{1}{z^2} + 10z = 0$ Multiply both sides by $z^2$ (since $z > 0$ , we're not multiplying by zero, so it's all good!): $-1 + 10z^3 = 0$ $10z^3 = 1$ $z^3 = \frac{1}{10}$ $z = \sqrt[3]{\frac{1}{10}}$

STEP 8

STEP 9

We already saw that $h(z)$ goes to infinity as $z$ gets close to zero or becomes very large.
This means there's no global maximum!

STEP 10

The global maximum of $h(z)$ does not exist.
The global minimum is $\sqrt[3]{10} + \sqrt[3]{\frac{5}{4}}$ .

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Studdy Solution

STEP 1

What is this asking? Find the biggest and smallest values of h(z)=1z+5z2h(z) = \frac{1}{z} + 5z^2h(z)=z1​+5z2 for positive zzz. Watch out! Don't forget to check what happens when zzz gets really close to zero and really, really big!

STEP 2

1. Explore the function's behavior2. Find the minimum3. Find the maximum

STEP 3

Let's see what happens when zzz gets super close to zero.As zzz approaches **zero**, 1z\frac{1}{z}z1​ becomes **gigantic**, so h(z)h(z)h(z) explodes to **infinity**!

STEP 4

Now, what happens when zzz becomes **huge**?Well, 5z25z^25z2 also becomes **massive**, so h(z)h(z)h(z) goes to **infinity** again!

STEP 5

To find the minimum, we need to find the **derivative** of h(z)h(z)h(z) and set it to **zero**!The derivative tells us how h(z)h(z)h(z) is changing.

STEP 6

ddzh(z)=ddz(1z+5z2)=−1z2+10z \frac{d}{dz} h(z) = \frac{d}{dz} \left( \frac{1}{z} + 5z^2 \right) = -\frac{1}{z^2} + 10z dzd​h(z)=dzd​(z1​+5z2)=−z21​+10z

STEP 7

STEP 8

STEP 9

We already saw that h(z)h(z)h(z) goes to **infinity** as zzz gets close to **zero** or becomes **very large**.This means there's no **global maximum**!

STEP 10

The global maximum of h(z)h(z)h(z) *does not exist*.The global minimum is 103+543\sqrt[3]{10} + \sqrt[3]{\frac{5}{4}}310​+345​​.

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What is this asking? Find the biggest and smallest values of $h(z) = \frac{1}{z} + 5z^2$ for positive $z$ . Watch out! Don't forget to check what happens when $z$ gets really close to zero and really, really big!

1. Explore the function's behavior
2. Find the minimum
3. Find the maximum

Let's see what happens when $z$ gets super close to zero.
As $z$ approaches zero, $\frac{1}{z}$ becomes gigantic, so $h(z)$ explodes to infinity!

Now, what happens when $z$ becomes huge?
Well, $5z^2$ also becomes massive, so $h(z)$ goes to infinity again!

To find the minimum, we need to find the derivative of $h(z)$ and set it to zero!
The derivative tells us how $h(z)$ is changing.

$\frac{d}{dz} h(z) = \frac{d}{dz} \left( \frac{1}{z} + 5z^2 \right) = -\frac{1}{z^2} + 10z$

We already saw that $h(z)$ goes to infinity as $z$ gets close to zero or becomes very large.
This means there's no global maximum!

The global maximum of $h(z)$ does not exist.
The global minimum is $\sqrt[3]{10} + \sqrt[3]{\frac{5}{4}}$ .