Concavity Analysis Of F(x) = X√(x² + 25) On [-5, 5]

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This article delves into the concavity of the function f(x) = x√(x² + 25) within the interval -5 ≤ x ≤ 5. Understanding concavity is crucial in calculus as it helps us determine the shape of a curve, specifically whether it curves upwards (concave up) or downwards (concave down). To achieve this, we'll embark on a step-by-step analysis, employing the power of differential calculus to reveal the intervals where the function exhibits these characteristics. Our journey will involve calculating the first and second derivatives of the function, meticulously identifying critical points, and subsequently constructing a sign chart to decipher the concavity across the specified interval. By the end of this exploration, we will have pinpointed the exact intervals where the function f(x) displays concave up and concave down behavior, providing a comprehensive understanding of its curvature.

Determining the First and Second Derivatives

The cornerstone of concavity analysis lies in the second derivative. It acts as a compass, guiding us toward regions of upward or downward curvature. Before we can ascertain the second derivative, we must first navigate the terrain of the first derivative. The first derivative, denoted as f'(x), unveils the rate of change of the function, which is crucial for finding critical points and understanding increasing and decreasing intervals. Let's embark on this calculation journey, employing the product rule and chain rule to unveil the first derivative of f(x) = x√(x² + 25).

Applying the product rule, where u = x and v = √(x² + 25):

  • u' = 1
  • v' = (1/2)(x² + 25)^(-1/2) * (2x) = x / √(x² + 25)

Thus, f'(x) = u'v + uv' = √(x² + 25) + x² / √(x² + 25). Now, we simplify this expression to obtain a more manageable form. By finding a common denominator, we get f'(x) = (x² + 25 + x²) / √(x² + 25) = (2x² + 25) / √(x² + 25). This first derivative will be instrumental in identifying critical points later on.

Now, let's venture into the realm of the second derivative, denoted as f''(x). This derivative reveals the rate of change of the slope, which directly corresponds to the concavity of the function. We'll employ the quotient rule to differentiate f'(x) = (2x² + 25) / √(x² + 25). Applying the quotient rule, where u = 2x² + 25 and v = √(x² + 25):

  • u' = 4x
  • v' = x / √(x² + 25) (as calculated previously)

The quotient rule states that f''(x) = (u'v - uv') / v². Plugging in our values, we have f''(x) = [4x√(x² + 25) - (2x² + 25)(x / √(x² + 25))] / (x² + 25). This expression looks complex, but with careful simplification, we can arrive at a more manageable form. Multiplying the numerator and denominator by √(x² + 25) to eliminate the nested fraction, we get f''(x) = [4x(x² + 25) - x(2x² + 25)] / (x² + 25)^(3/2). Further simplification yields f''(x) = (4x³ + 100x - 2x³ - 25x) / (x² + 25)^(3/2) = (2x³ + 75x) / (x² + 25)^(3/2). This simplified second derivative is the key to unraveling the concavity of f(x).

Identifying Points of Inflection

Points of inflection are the pivotal locations where a function's concavity undergoes a transformation – a shift from curving upwards (concave up) to curving downwards (concave down), or vice versa. These transition points are identified where the second derivative, f''(x), either equals zero or is undefined. To pinpoint these crucial points for our function, f(x) = x√(x² + 25), we'll focus on analyzing the second derivative we previously derived: f''(x) = (2x³ + 75x) / (x² + 25)^(3/2).

The denominator, (x² + 25)^(3/2), is always positive for any real value of x because x² is non-negative, and adding 25 ensures a positive result. Raising this positive quantity to the power of 3/2 maintains its positivity. Therefore, the second derivative f''(x) is undefined.

Next, we turn our attention to the numerator, 2x³ + 75x. We seek values of x that make this expression equal to zero. Setting 2x³ + 75x = 0, we can factor out an x, resulting in x(2x² + 75) = 0. This equation holds true when either x = 0 or 2x² + 75 = 0. Let's examine the second factor, 2x² + 75 = 0. This leads to 2x² = -75, and subsequently, x² = -75/2. Since the square of any real number cannot be negative, there are no real solutions arising from this factor. Consequently, the only point where the second derivative equals zero is at x = 0.

Thus, x = 0 stands as the sole potential point of inflection for our function within the interval [-5, 5]. This point divides our interval into two subintervals: [-5, 0) and (0, 5]. We will now investigate the sign of the second derivative within each of these subintervals to determine the concavity of f(x) in those regions.

Determining Intervals of Concavity

To definitively establish the intervals where the function f(x) = x√(x² + 25) exhibits concave up or concave down behavior, we must now examine the sign of the second derivative, f''(x) = (2x³ + 75x) / (x² + 25)^(3/2), across the intervals defined by our potential inflection point, x = 0. Recall that the denominator, (x² + 25)^(3/2), is always positive, so the sign of f''(x) is solely determined by the numerator, 2x³ + 75x.

We will analyze the sign of f''(x) on the intervals [-5, 0) and (0, 5] by choosing test values within each interval and evaluating the numerator.

Interval [-5, 0):

Let's select a test value, say x = -1. Plugging this into the numerator, we get 2(-1)³ + 75(-1) = -2 - 75 = -77. Since this is negative, f''(x) is negative on the interval [-5, 0). A negative second derivative indicates that the function is concave down on this interval.

Interval (0, 5]:

Now, let's choose a test value from this interval, such as x = 1. Substituting this into the numerator, we obtain 2(1)³ + 75(1) = 2 + 75 = 77. This positive result signifies that f''(x) is positive on the interval (0, 5]. Consequently, the function is concave up on this interval.

In summary, our analysis reveals that the function f(x) = x√(x² + 25) is concave down on the interval [-5, 0) and concave up on the interval (0, 5]. The point x = 0 marks the transition between these concavity states, solidifying its role as a point of inflection.

In conclusion, by meticulously applying differential calculus techniques, we've successfully dissected the concavity of the function f(x) = x√(x² + 25) over the interval -5 ≤ x ≤ 5. Our journey began with the calculation of the first and second derivatives, paving the way for the identification of the potential inflection point at x = 0. Through careful analysis of the sign of the second derivative on the intervals [-5, 0) and (0, 5], we definitively established that the function is concave down on the interval [-5, 0) and concave up on the interval (0, 5]. This comprehensive exploration underscores the power of calculus in unraveling the intricate behavior of functions, providing valuable insights into their curvature and overall shape.