Suppose Is A Twice Differentiable Function With The Known Va

Suppose Ht Is A Twice Differentiable Function With The Known Values

Suppose h(t) is a twice-differentiable function with the known values: h(t), h'(t), h''(t). Justifying your reasoning, answer the following:

a) Determine where h(t) = 0 on the interval (0, 4).

b) Determine where h'(t) = 0 on the interval (0, 4).

c) Determine where h''(t) = 0 on the interval (0, 4).

2. Use limits to find and verify all asymptotes in the graph of √(5 + x²) / (3 - x).

Paper For Above instruction

Analyzing the behavior of functions and their derivatives over specific intervals is fundamental in calculus, especially when determining points of interest such as zeros, critical points, inflection points, and asymptotes. The given problem involves a twice-differentiable function h(t), with known values, and requires a detailed analysis of its zeros and the zeros of its derivatives over the interval (0, 4). Additionally, the problem extends to examining the asymptotic behavior of the function f(x) = √(5 + x²) / (3 - x), particularly through limits to identify asymptotes.

Part A: Determining where h(t) = 0 in (0, 4)

The first step is to analyze the values of h(t) within the interval (0, 4). Since the problem states that h(t) is known at specific points and derivatives, but does not specify the actual values, the general approach involves using the Intermediate Value Theorem (IVT). If h(t) is continuous over [a, b] within (0, 4), and if h(a) and h(b) have opposite signs (i.e., h(a) * h(b)

For example, suppose h(0) > 0 and h(4)

Part B: Determining where h'(t) = 0 in (0, 4)

Since h(t) is twice differentiable, h'(t) is continuous. Critical points occur where h'(t) = 0, which might correspond to local maxima, minima, or inflection points depending on the behavior of h''(t). To find these critical points, one typically employs the first derivative test or examines the sign of h'(t) around points where h'(t) could be zero.

Using known values or the derivatives' behavior, possibly obtained from the given data, one can identify intervals where h'(t) changes sign, indicating a zero of h'(t). For example, if h'(t) changes from positive to negative at some t in (0, 4), then h(t) has a local maximum at that point. Conversely, a change from negative to positive indicates a local minimum. Continuity ensures that zeros of h'(t) can be detected through sign changes across the interval.

Part C: Determining where h''(t) = 0 in (0, 4)

The second derivative h''(t) signifies concavity and inflection points. Zeros of h''(t) indicate potential inflection points. To find and verify these, one must analyze the sign of h''(t) and its behavior over (0, 4). If h''(t) changes sign around some t, then h(t) has an inflection point there. Given the known behavior or sample values, sign charts or the second derivative test can confirm these points.

Part 2: Asymptotic Behavior of the Function

Consider the function f(x) = √(5 + x²) / (3 - x). Asymptotes are lines that the graph approaches but never touches. To find vertical asymptotes, we examine the points where the denominator approaches zero, i.e., where 3 - x = 0, which implies x = 3. Therefore, there is a vertical asymptote at x = 3.

To verify this, evaluate the limits approaching x = 3 from the left and right:

\[

\lim_{x \to 3^-} f(x) \quad \text{and} \quad \lim_{x \to 3^+} f(x)

\]

As x approaches 3 from the left, 3 - x approaches zero positively, making the denominator tend to zero. Since √(5 + x²) remains finite and positive near x = 3, f(x) tends toward +∞ or -∞ depending on the numerator's behavior. Similarly, for x approaching 3 from the right, the denominator tends to zero negatively, and the function tends toward ∓∞, confirming the vertical asymptote.

Horizontal asymptotes are determined by examining the limits of f(x) as x approaches ∞ and -∞:

\[

\lim_{x \to \infty} f(x) \quad \text{and} \quad \lim_{x \to -\infty} f(x)

\]

For large |x|, √(5 + x²) behaves like |x|, so:

\[

f(x) \sim \frac{|x|}{3 - x}

\]

As x → ∞, the dominant term is -x in the denominator and |x| in the numerator, so f(x) approaches -1. As x → -∞, |x| = -x, so similarly, the function tends toward -1. Therefore, the horizontal asymptote is y = -1.

Conclusion

In conclusion, analyzing the zeros of h(t), h'(t), and h''(t) over the interval (0, 4) involves applying the Intermediate Value Theorem and understanding the behavior of derivatives. The asymptotic analysis of the function f(x) = √(5 + x²) / (3 - x) reveals a vertical asymptote at x = 3 and a horizontal asymptote at y = -1, identified through limits. These analyses are crucial for sketching the graph and understanding the geometric behavior of the functions involved.

References

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