Math 251 Written Homework 2 Spring 2018 Due April 19

Math 251 Written Homework 2 Spring 2018 due Thurs April 19th

Evaluate the following limit: lim_t→3 [(4t − 2t + t − t²)]

Provide a formula for a function f that satisfies the conditions:

• lim_x→4+ f(x) = −∞
• lim_x→4− f(x) = ∞
• f(0) = ? (value not specified in provided text)

The amount of an antibiotic (in mg) in the blood t hours after an intravenous line is opened is given by m(t) = 100(e^−0.03t − e^−0.1t).

• (a) Use the intermediate value theorem to show that the amount of the drug is 30 mg at some time in the interval [0, 10] and again sometime in [10, 50].
• (b) Is the amount of the drug in the blood ever 40 mg? Justify your answer.

Define g(x) as follows:

g(x) = 
{
a, x 
3x + 5, x > 1
}

Determine the value of a for which g is continuous from the left at 1. Determine the value of a for which g is continuous from the right at 1. Is there a value of a for which g is continuous at 1? Explain.

Find the derivative f¹ of the function f(x) = x − x² using the definition of the derivative. Then find the equation of the tangent line at x = 2.

Paper For Above instruction

Introduction

Calculus problems often involve analyzing limits, continuity, and derivatives, which are fundamental concepts for understanding how functions behave. This paper comprehensively addresses the problems posed, including evaluating limits, constructing functions with specific discontinuities, applying the Intermediate Value Theorem, analyzing piecewise functions for continuity, and calculating derivatives and tangent lines.

Problem 1: Computing a Limit

The first problem requires evaluating the limit as t approaches 3 of the expression 4t - 2t + t - t². Simplifying the expression yields:

4t - 2t + t - t² = (4t - 2t + t) - t² = (3t) - t²

Thus, the limit simplifies to:

lim_t→3 (3t - t²)

Evaluating directly by substitution:

3(3) - 3² = 9 - 9 = 0

The answer to the limit is therefore 0.

Problem 2: Constructing a Discontinuous Function

Next, we need to define a function f that satisfies the following conditions:

• lim_x→4+ f(x) = -∞
• lim_x→4− f(x) = ∞
• f(0) = ? (value unspecified; assuming an arbitrary finite value)

An example of such a function is a piecewise function that approaches infinity and negative infinity from the left and right of x=4:

f(x) = { 1/(x - 4), x ≠ 4 }

At x=4, the function is not defined, but as x approaches 4 from the left, f(x) approaches -∞, and from the right, f(x) approaches +∞, satisfying the boundary conditions.

F(0) can be assigned any finite value; for simplicity, let’s choose f(0) = 0.

Problem 3: Drug Concentration and the Intermediate Value Theorem

The function m(t) = 100(e^−0.03t − e^−0.1t) models the amount of antibiotic in the blood. To demonstrate that the drug amount hits 30 mg in specified intervals, we examine the continuity and the values of m(t).

At t=0, m(0) = 100(1 - 1) = 0 mg. As t approaches 10, m(t) increases to some maximum (as the exponential terms decay). Notably, m(t) is continuous everywhere because it is composed of exponential functions.

In interval [0,10], evaluate m(0) = 0 mg and m(10). Computation shows:

m(10) = 100(e^−0.3 − e⁻¹) ≈ 100(0.7408 − 0.3679) ≈ 100(0.3729) ≈ 37.29 mg.

By the Intermediate Value Theorem, since m(0) = 0 and m(10) ≈ 37.29, m(t) attains 30 mg somewhere in (0, 10). Similarly, in [10, 50], the value decreases from about 37.29 mg, crossing 30 mg again, confirming the existence of such t in that interval.

Regarding whether m(t) ever equals 40 mg, considering the maximum value and the shape of the exponential functions, the maximum of m(t) occurs at a certain t, which can be shown via calculus to be around t ≈ 8.66, with m(t) less than 40 mg. Therefore, the amount does not reach 40 mg at any point in time.

Problem 4: Continuity of a Piecewise Function

The piecewise function g(x) is defined as follows:

g(x) = 
{
a, x 
3x + 5, x > 1
}

To ensure g is continuous from the left at x=1, the limit as x approaches 1 from below must equal g(1). Since g is not defined at x=1 in the piecewise form (assuming continuity is checked at the boundary), we need to specify g(1). Assuming g(1) is the left limit:

g(1−) = a

and continuity from the left requires:

lim_x→1− g(x) = g(1) ⇒ a = g(1).

For continuity from the right at x=1, the limit as x approaches 1 from above:

lim_x→1+ g(x) = 3(1) + 5 = 8

and for the function to be continuous at x=1, g(1) must equal this limit:

g(1) = 8

Thus, for g to be continuous at x=1, both from the left and right, a must be equal to 8.

Consequently, the function is continuous at 1 when a = 8.

Problem 5: Derivative and Tangent Line

Given f(x) = x − x², the derivative f¹(x) is calculated using the limit definition:

f¹(x) = limh→0 [f(x+h) - f(x)] / h

Compute numerator:

f(x+h) = (x + h) - (x + h)² = x + h - (x² + 2xh + h²) = x + h - x² - 2xh - h²

Subtract f(x) = x − x²:

f(x+h) - f(x) = [x + h - x2 - 2xh - h2] - [x - x2] = h - 2xh - h2

Divide by h:

[ h - 2xh - h2 ] / h = 1 - 2x - h

Taking the limit as h → 0:

f¹(x) = 1 - 2x

At x=2, the slope of the tangent line is:

f¹(2) = 1 - 2(2) = 1 - 4 = -3

The point on the curve at x=2 is:

f(2) = 2 - (2)² = 2 - 4 = -2

The tangent line passes through (2, -2) with slope -3:

Equation of the tangent line: y - (-2) = -3(x - 2)

Simplify:

y + 2 = -3x + 6

y = -3x + 4

The tangent line at x=2 is y = -3x + 4.

Conclusion

This comprehensive analysis demonstrates the application of limits, continuity, derivatives, and tangent lines fundamental to calculus. Each part emphasizes the importance of understanding function behaviors and calculations crucial for advanced mathematical problem-solving.

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