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Chapter 4: Problem 37

\(23-46\) Find \(f\) $$f^{\prime \prime}(x)=24 x^{2}+2 x+10, f(1)=5, \quad f^{\prime}(1)=-3$$

Short Answer

Expert verified
\(f(x) = 2x^4 + \frac{x^3}{3} + 5x^2 - 22x + \frac{14}{3}\).

Step by step solution

01

Integrate the Second Derivative

We start by finding the first derivative, which requires us to integrate the second derivative function. The second derivative given is \( f''(x) = 24x^2 + 2x + 10 \). Integrating term by term, we get:\[ f'(x) = \int (24x^2 + 2x + 10) \, dx = 8x^3 + x^2 + 10x + C_1 \] where \( C_1 \) is the constant of integration.
02

Use Initial Condition for f'(x)

We use the initial condition \( f'(1) = -3 \) to find the constant \( C_1 \). Plugging \( x = 1 \) into the expression for \( f'(x) \), we have:\[ -3 = 8(1)^3 + (1)^2 + 10(1) + C_1 \]Simplifying gives:\[ -3 = 8 + 1 + 10 + C_1 \]\[ -3 = 19 + C_1 \]\[ C_1 = -22 \]
03

Integrate the First Derivative

Now with \( C_1 \) found, we integrate \( f'(x) \) to find \( f(x) \). We have:\[ f'(x) = 8x^3 + x^2 + 10x - 22 \]Integrating term by term:\[ f(x) = \int (8x^3 + x^2 + 10x - 22) \, dx = 2x^4 + \frac{x^3}{3} + 5x^2 - 22x + C_2 \] where \( C_2 \) is another constant of integration.
04

Use Initial Condition for f(x)

Finally, we use the initial condition \( f(1) = 5 \) to find \( C_2 \). Substituting \( x = 1 \) into \( f(x) \) gives:\[ 5 = 2(1)^4 + \frac{1^3}{3} + 5(1)^2 - 22(1) + C_2 \]Simplifying this:\[ 5 = 2 + \frac{1}{3} + 5 - 22 + C_2 \]\[ 5 = -15 + \frac{1}{3} + C_2 \]\[ \frac{14}{3} = C_2 \]
05

Write the Final Expression for f(x)

Now we have all the components to write the full expression for \( f(x) \):\[ f(x) = 2x^4 + \frac{x^3}{3} + 5x^2 - 22x + \frac{14}{3} \]This function satisfies the given initial conditions.

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Key Concepts

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

Integrating Polynomial Functions
When we talk about integrating polynomial functions, we're essentially reversing the process of differentiation. For example, if we want to find a function's first derivative from its second derivative, we integrate the second derivative.
\[ f''(x) = 24x^2 + 2x + 10 \]
To go from this second derivative to the first, we perform integration term by term.
This means:
  • The integral of \(24x^2\) is \(8x^3 + C\), because the power of \(x\) increases by 1 and we divide by the new power (3).
  • The integral of \(2x\) becomes \(x^2\), because similarly, the power of \(x\) increases by 1 and we divide by this new power (2).
  • Finally, the integral of a constant \(10\) is \(10x\).
These terms combine with a constant of integration, which we'll discuss in more detail later.
Initial Conditions in Calculus
In calculus, initial conditions allow us to find specific solutions to differential equations. When you integrate, you always add a constant of integration because differentiation loses this constant.
Using an initial condition, like \( f'(1) = -3 \), helps us determine this constant of integration.
In our example, we substitute \(x = 1\) into our derived equation for \(f'(x)\). This calculation allowed us to solve for the constant \(C_1\), determining a specific form of the function.
  • The equation becomes \(-3 = 8(1)^3 + (1)^2 + 10(1) + C_1\).
  • Simplifying gives \(C_1 = -22\).
With this initial condition, we have pinned down the particular first derivative that fits the scenario described in the problem.
Constant of Integration in Calculus
The constant of integration, often represented as \(C\), arises whenever we integrate a function. This constant is crucial because it accounts for any vertical shift in the antiderivative.
Each time we integrate a function, we add a new constant of integration. For instance, after finding \(f'(x)\), integrating again to find \(f(x)\) introduces another constant \(C_2\).
Given \(f(1) = 5\), we can determine this \(C_2\).
  • Substitute \(x = 1\) into \( f(x) = 2x^4 + \frac{x^3}{3} + 5x^2 - 22x + C_2 \).
  • Simplify the equation to solve for \(C_2\), resulting in \(\frac{14}{3}\).
The constants of integration make sure that our final function perfectly matches the conditions provided, leaving us with the complete polynomial function.

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Most popular questions from this chapter

since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. A rain-drop has an initial downward velocity of 10 \(\mathrm{m} / \mathrm{s}\) and its downward acceleration is $$a=\left\\{\begin{array}{ll}{9-0.9 t} & {\text { if } 0 \leq t \leq 10} \\\ {0} & {\text { if } t>10}\end{array}\right.$$ If the raindrop is initially 500 \(\mathrm{m}\) above the ground, how long does it take to fall?

A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t) = \(60 t,\) at which time the fuel is exhausted and it becomes a freely "falling" body. Fourteen seconds later, the rocket's parachute opens, and the (downward) velocity slows linearly to \(-18 \mathrm{ft} / \mathrm{s}\) in 5 \(\mathrm{s}\) . The rocket then "floats" to the ground at that rate. (a) Determine the position function s and the velocity function \(v\) (for all times t). Sketch the graphs of s and v. (b) At what time does the rocket reach its maximum height, and what is that height? (c) At what time does the rocket land?

A high-speed bullet train accelerates and decelerates at the rate of 4 \(\mathrm{ft} / \mathrm{s}^{2} .\) Its maximum cruising speed is 90 \(\mathrm{mi} / \mathrm{h}\) . (a) What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes? (b) Suppose that the train starts from rest and must come to a complete stop in 15 minutes. What is the maximum distance it can travel under these conditions? (c) Find the minimum time that the train takes to travel between two consecutive stations that are 45 miles apart. (d) The trip from one station to the next takes 37.5 minutes. How far apart are the stations?

Produce graphs of \(f\) that reveal all the important aspects of the curve. In particular, you should use graphs of \(f^{\prime}\) and \(f "\) to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points. \(f(x)=x^{6}-10 x^{5}-400 x^{4}+2500 x^{3}\)

\(17-22\) Use Newton's method to find all roots of the equation cor- rect to six decimal places. \((x-2)^{2}=\ln x\)
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