We need to find the lower and upper bounds of the function:

$f(x) = \sqrt{1 + x^4}$

on the interval $[0,1]$.

Step 1: Determine the Lower Bound

The lower bound is the minimum value of $f(x)$ on the given interval.

• Since $x^4$ is always non-negative, the minimum value occurs when $x$ is smallest, which is at $x = 0$.

• Plugging $x = 0$ into the function:

  $f(0) = \sqrt{1 + 0^4} = \sqrt{1} = 1.$

Thus, the lower bound is $1$.

Step 2: Determine the Upper Bound

The upper bound is the maximum value of $f(x)$ on the given interval.

• The function $f(x) = \sqrt{1 + x^4}$ is increasing because $x^4$ increases as $x$ increases.

• The maximum value occurs at $x = 1$:

  $f(1) = \sqrt{1 + 1^4} = \sqrt{1 + 1} = \sqrt{2}.$

Thus, the upper bound is $\sqrt{2}$.

Final Answer

The function is bounded between 1 and $\sqrt{2}$ on $[0,1]$:

$1 \leq f(x) \leq \sqrt{2}.$

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Would you like a step-by-step explanation for any part?

Here are 5 related questions to explore:

1. How do we determine whether a function is increasing or decreasing?
2. Can we use derivatives to confirm the bounds of $f(x)$?
3. How do we find absolute extrema of a function on a closed interval?
4. What happens if the interval changes to $[-1,1]$?
5. How can we approximate $\sqrt{2}$ to better understand the upper bound?

Tip: When working with square root functions, always check whether the expression inside the square root is increasing or decreasing to determine bounds effectively. 🚀