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Determine where the function is concave upward and where it is concave downward. $$ f(x)=\sqrt{4-x} $$

Short Answer

Expert verified
The function \(f(x) = \sqrt{4-x}\) is concave upward on the interval x ∈ \((-\infty, 4)\) and has no intervals where it is concave downward.

Step by step solution

01

Find the first derivative

To find the first derivative, we will differentiate the given function f(x). The function is given as \[f(x) = \sqrt{4-x}\] We can rewrite this as: \[f(x) = (4-x)^\frac{1}{2}\] Now find the first derivative f'(x) using the power rule and chain rule: \[f'(x) = \frac{1}{2}(4-x)^{-\frac{1}{2}}(-1) = - \frac{1}{2}(4-x)^{-\frac{1}{2}}\]
02

Find the second derivative

To find the second derivative, we will differentiate the first derivative f'(x) with respect to x. The first derivative f'(x) is given as: \[f'(x) = -\frac{1}{2}(4-x)^{-\frac{1}{2}}\] Now differentiate f'(x) with respect to x using the power rule and chain rule: \[f''(x) = -\frac{1}{2}(-\frac{1}{2})(4-x)^{-\frac{3}{2}}(-1) = \frac{1}{4}(4-x)^{-\frac{3}{2}}\]
03

Determine where f''(x) is positive or negative

Now that we have the second derivative f''(x), we can determine where it is positive or negative. Consider the second derivative: \[f''(x) = \frac{1}{4}(4-x)^{-\frac{3}{2}}\] First, notice that the second derivative is defined only for x in the interval \((-\infty, 4)\). Now, since the term \((4-x)^{-\frac{3}{2}}\) will always be positive (as it has an even exponent and 4 > x), the entire expression will be positive as well. So, f''(x) is positive for the interval x ∈ \((-\infty, 4)\). Since there is no value of x for which f''(x) is negative or zero, the function has no concave downward intervals.
04

Write the final answer

Using our analysis of the second derivative f''(x), we can write the final answer: The function \(f(x) = \sqrt{4-x}\) is concave upward on the interval x ∈ \((-\infty, 4)\). There are no intervals where the function is concave downward.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second Derivative Test
Understanding the concavity of a function involves using the second derivative test, which is a reliable method for determining whether a point on a graph is a local maximum or minimum. But before we get to extremes, let's focus on concavity. A function's concavity refers to the direction in which the curve opens.

When you calculate the second derivative, denoted as \(f''(x)\), and find that it is positive over an interval, the function is concave upward on that interval. Similarly, if \(f''(x)\) is negative, the function is concave downward. An easy way to remember this: Positive is like a smile (\(\cup\)) — concave upward, and negative is like a frown (\(\cap\)) — concave downward.

For the given function \(f(x) = \sqrt{4-x}\), the second derivative \(f''(x)\) turns out to be positive across its entire domain, so we say that the function is concave upward everywhere within that domain. Note that the test fails if \(f''(x)\) is zero, as this indicates a possible inflection point, which would require further investigation.
Concave Upward
When a function's graph is shaped like a cup, \(\cup\), and opens upwards, it's described as concave upward. Imagine holding a ball on this curve: the ball would naturally roll to the bottom, indicating that at any point on the curve, the tangent line lies below the curve. This curvature signifies that the rate of change of the function's slope is increasing.

In terms of the second derivative, this happens when \(f''(x) > 0\). In the example of \(f(x) = \sqrt{4-x}\), since the function’s second derivative is always positive for x in \((-\infty, 4)\), it means that the tangent line at any point in this interval would lie below the curve — a classic sign of a concave upward graph.
Concave Downward
Contrarily, a function is concave downward when its graph creates a shape like an upside-down cup, \(\cap\). If you placed a ball on this curve, it would roll away from the center, signifying that the tangent line at any point on the curve lies above the curve. Here, the function's slope increases at a decreasing rate.

A negative second derivative, \(f''(x) < 0\), indicates a concave downward interval. However, for our function \(f(x) = \sqrt{4-x}\), the second derivative is never negative, which means the graph never has the concave downward characteristic within its domain. Understanding this concept is vital, as it affects the behavior of the graph and can help identify the type of extremum — maximum or minimum — at a given point, if any.

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Most popular questions from this chapter

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