Find F(x) From Its Derivative & A Point On The Curve

by Andrew McMorgan 53 views

Hey Plastik Magazine readers! Let's dive into a calculus problem where we'll find a function given its derivative and a point on the curve. This kind of problem is super common in calculus, and mastering it will definitely level up your math skills. We'll break it down step-by-step, so you can follow along easily. So, buckle up, and let's get started!

Understanding the Problem

Before we jump into solving, let's quickly recap what we have. We're given the derivative of a function, f′(x)=(2−x2)3x2{ f'(x) = \frac{(2-x^2)^3}{x^2} }, and we need to find the original function, f(x){ f(x) }. Plus, we know that the curve y=f(x){ y = f(x) } passes through the point (−2,9){ (-2, 9) }. This point is crucial because it will help us find the constant of integration after we integrate f′(x){ f'(x) }.

Our main objective here is twofold:

  1. Show that f′(x){ f'(x) } can be expressed in the form 8x−2−12+Ax2+Bx4{ 8x^{-2} - 12 + Ax^2 + Bx^4 } and find the values of the constants A{ A } and B{ B }.
  2. Find the actual function f(x){ f(x) } using the given point (−2,9){ (-2, 9) }.

Expanding and Simplifying f'(x)

First, we need to expand and simplify the given derivative, f′(x)=(2−x2)3x2{ f'(x) = \frac{(2-x^2)^3}{x^2} }. This involves expanding the cube in the numerator and then dividing each term by x2{ x^2 }. Let's break it down:

(2−x2)3=(2−x2)(2−x2)(2−x2){ (2-x^2)^3 = (2-x^2)(2-x^2)(2-x^2) }

Let's first multiply the first two factors:

(2−x2)(2−x2)=4−4x2+x4{ (2-x^2)(2-x^2) = 4 - 4x^2 + x^4 }

Now, multiply this result by the third factor (2−x2){ (2-x^2) }:

(4−4x2+x4)(2−x2)=8−8x2+2x4−4x2+4x4−x6{ (4 - 4x^2 + x^4)(2-x^2) = 8 - 8x^2 + 2x^4 - 4x^2 + 4x^4 - x^6 }

Combine like terms:

8−12x2+6x4−x6{ 8 - 12x^2 + 6x^4 - x^6 }

So, (2−x2)3=8−12x2+6x4−x6{ (2-x^2)^3 = 8 - 12x^2 + 6x^4 - x^6 }. Now we divide this by x2{ x^2 } to get f′(x){ f'(x) }:

f′(x)=8−12x2+6x4−x6x2{ f'(x) = \frac{8 - 12x^2 + 6x^4 - x^6}{x^2} }

Divide each term by x2{ x^2 }:

f′(x)=8x2−12x2x2+6x4x2−x6x2{ f'(x) = \frac{8}{x^2} - \frac{12x^2}{x^2} + \frac{6x^4}{x^2} - \frac{x^6}{x^2} }

Simplify each term:

f′(x)=8x−2−12+6x2−x4{ f'(x) = 8x^{-2} - 12 + 6x^2 - x^4 }

Comparing this with the given form f′(x)=8x−2−12+Ax2+Bx4{ f'(x) = 8x^{-2} - 12 + Ax^2 + Bx^4 }, we can easily identify the constants:

A=6{ A = 6 } and B=−1{ B = -1 }

So, we've successfully shown that f′(x)=8x−2−12+6x2−x4{ f'(x) = 8x^{-2} - 12 + 6x^2 - x^4 }.

Finding the Function f(x)

Now that we have f′(x){ f'(x) } in a simplified form, we can integrate it to find f(x){ f(x) }. Remember, integration is the reverse process of differentiation. We'll integrate each term of f′(x){ f'(x) } with respect to x{ x }:

f′(x)=8x−2−12+6x2−x4{ f'(x) = 8x^{-2} - 12 + 6x^2 - x^4 }

Integrate each term:

∫8x−2dx=8∫x−2dx=8⋅x−1−1=−8x−1{ \int 8x^{-2} dx = 8 \int x^{-2} dx = 8 \cdot \frac{x^{-1}}{-1} = -8x^{-1} }

∫−12dx=−12x{ \int -12 dx = -12x }

∫6x2dx=6∫x2dx=6⋅x33=2x3{ \int 6x^2 dx = 6 \int x^2 dx = 6 \cdot \frac{x^3}{3} = 2x^3 }

∫−x4dx=−∫x4dx=−x55{ \int -x^4 dx = - \int x^4 dx = - \frac{x^5}{5} }

So, integrating f′(x){ f'(x) } gives us:

f(x)=−8x−1−12x+2x3−x55+C{ f(x) = -8x^{-1} - 12x + 2x^3 - \frac{x^5}{5} + C }

Here, C{ C } is the constant of integration. We need to find this constant using the given point (−2,9){ (-2, 9) }. This point tells us that when x=−2{ x = -2 }, f(x)=9{ f(x) = 9 }. Let's plug these values into our equation:

9=−8(−2)−1−12(−2)+2(−2)3−(−2)55+C{ 9 = -8(-2)^{-1} - 12(-2) + 2(-2)^3 - \frac{(-2)^5}{5} + C }

Simplify each term:

9=−8(−12)+24+2(−8)−−325+C{ 9 = -8(-\frac{1}{2}) + 24 + 2(-8) - \frac{-32}{5} + C }

9=4+24−16+325+C{ 9 = 4 + 24 - 16 + \frac{32}{5} + C }

9=12+325+C{ 9 = 12 + \frac{32}{5} + C }

Now, let's isolate C{ C }:

C=9−12−325{ C = 9 - 12 - \frac{32}{5} }

C=−3−325{ C = -3 - \frac{32}{5} }

Convert -3 to a fraction with a denominator of 5:

C=−155−325{ C = -\frac{15}{5} - \frac{32}{5} }

C=−475{ C = -\frac{47}{5} }

Now that we have the constant of integration, we can write the complete function f(x){ f(x) }:

f(x)=−8x−1−12x+2x3−x55−475{ f(x) = -8x^{-1} - 12x + 2x^3 - \frac{x^5}{5} - \frac{47}{5} }

Summarizing the Solution

Alright, guys, we've tackled this calculus problem like pros! Let's recap what we've done:

  1. We started with the derivative f′(x)=(2−x2)3x2{ f'(x) = \frac{(2-x^2)^3}{x^2} }.
  2. We showed that f′(x){ f'(x) } can be expressed as 8x−2−12+Ax2+Bx4{ 8x^{-2} - 12 + Ax^2 + Bx^4 } and found that A=6{ A = 6 } and B=−1{ B = -1 }.
  3. We integrated f′(x){ f'(x) } to find f(x)=−8x−1−12x+2x3−x55+C{ f(x) = -8x^{-1} - 12x + 2x^3 - \frac{x^5}{5} + C }.
  4. We used the point (−2,9){ (-2, 9) } to find the constant of integration, C=−475{ C = -\frac{47}{5} }.
  5. Finally, we wrote the complete function: f(x)=−8x−1−12x+2x3−x55−475{ f(x) = -8x^{-1} - 12x + 2x^3 - \frac{x^5}{5} - \frac{47}{5} }.

Importance of Showing Steps

In mathematics, especially in calculus, showing your steps is super important. It's not just about getting the final answer; it's about demonstrating that you understand the process. When you show each step, you're more likely to catch any mistakes you might make along the way. Plus, it helps anyone who's reading your work to follow your logic and understand how you arrived at the solution. Trust us, clear and detailed steps are your best friend in math!

Common Mistakes to Avoid

When solving problems like this, there are a few common pitfalls to watch out for:

  • Incorrectly Expanding the Cube: Expanding (2−x2)3{ (2-x^2)^3 } requires careful multiplication. A small mistake here can throw off the entire solution. Always double-check your expansion.
  • Forgetting the Constant of Integration: When you integrate, don't forget to add the constant of integration, C{ C }. This constant is crucial for finding the specific function given a point on the curve.
  • Making Sign Errors: Sign errors are super common in calculus. Pay close attention to the signs of each term when integrating and substituting values.
  • Incorrectly Integrating Terms: Make sure you apply the power rule of integration correctly. For example, ∫xndx=xn+1n+1+C{ \int x^n dx = \frac{x^{n+1}}{n+1} + C }, where n≠−1{ n \neq -1 }.

Tips for Mastering Calculus Problems

Want to become a calculus whiz? Here are some tips to help you master these types of problems:

  • Practice Regularly: The more you practice, the more comfortable you'll become with the concepts and techniques. Try solving similar problems from your textbook or online resources.
  • Understand the Concepts: Don't just memorize formulas; understand the underlying concepts. This will help you apply them in different situations.
  • Break Down Problems: Complex problems can seem daunting, but if you break them down into smaller steps, they become much more manageable.
  • Check Your Work: Always check your work, especially on exams. Make sure you haven't made any simple mistakes.
  • Seek Help When Needed: If you're struggling with a particular concept or problem, don't hesitate to ask for help from your teacher, classmates, or online forums.

Final Thoughts

So, there you have it! We've successfully found f(x){ f(x) } from its derivative and a point on the curve. These types of problems might seem tough at first, but with practice and a solid understanding of the fundamentals, you'll be acing them in no time. Keep practicing, stay curious, and remember to show your steps! You've got this, Plastik Magazine readers! Keep shining, and catch you in the next one!