Exercise 7.1 · Q22 (MCQ): $f'(x) = 4x^3 - \frac{3}{x^4}$⁠, $f(2) = 0$

Basics you need

Solution

1. Write the derivative as powers of $x$

   $$f'(x) = 4x^3 - \frac{3}{x^4} = 4x^3 - 3x^{-4}, \qquad x \ne 0.$$
   The power formula needs single powers $x^n$⁠. A power in the denominator moves up with its exponent's sign changed. The term $\frac{3}{x^4}$ is undefined at $x = 0$⁠, so we work with $x \ne 0$⁠.
   $\frac{1}{x^n} = x^{-n}$⁠. Basics: powers of x

2. Integrate to get the general antiderivative

   $$\begin{aligned} 4\int x^3\,dx &= 4\cdot\frac{x^4}{4} = x^4, \\ -3\int x^{-4}\,dx &= -3\cdot\frac{x^{-3}}{-3} = x^{-3}. \end{aligned}$$

   So

   $$f(x) = x^4 + \frac{1}{x^3} + C.$$
   Power formula with $n = 3$ (new power 4) and $n = -4$ (new power $-3$⁠); neither is $-1$⁠. In the second term the two minus signs cancel: $-3 \div (-3) = 1$⁠. Property V allows the constants 4 and $-3$ to come out. Because $f$ is some antiderivative of $f'$⁠, it equals this expression for one particular value of $C$⁠, which we now find.
   Formula (i), Section 7.2; Property V, Section 7.2.1. Lesson: integration as anti-differentiation

3. Use $f(2) = 0$ to find $C$

   $$\begin{aligned} f(2) &= 2^4 + \frac{1}{2^3} + C \\ &= 16 + \frac18 + C = \frac{129}{8} + C. \end{aligned}$$

   Setting $f(2) = 0$⁠:

   $$\frac{129}{8} + C = 0 \quad\Longrightarrow\quad C = -\frac{129}{8}.$$

   Hence

   $$f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}.$$
   The condition is one equation in the one unknown $C$⁠, so it determines $C$ uniquely. $16 = \frac{128}{8}$⁠, and $\frac{128}{8} + \frac18 = \frac{129}{8}$⁠. Geometrically, of all the curves $y = x^4 + \frac{1}{x^3} + C$ (vertical shifts of each other), exactly one passes through $(2, 0)$⁠; see the figure. Strictly, $f'$ is given on two separate intervals, $x < 0$ and $x > 0$⁠, and $f(2) = 0$ fixes the constant only on $x > 0$⁠. The options each use one formula for all $x \ne 0$⁠, and with that (usual) convention the answer is as above.
   This is exactly the method of Example 4, Section 7.2.1, where $F(0) = 3$ gives $C = 3$⁠. Basics: integration undoes differentiation

4. Check: differentiate and substitute

   $$\begin{aligned} \frac{d}{dx}\left(x^4 + x^{-3} - \tfrac{129}{8}\right) &= 4x^3 - 3x^{-4} \\ &= 4x^3 - \frac{3}{x^4}, \\ f(2) = 16 + \tfrac18 - \tfrac{129}{8} &= \tfrac{128 + 1 - 129}{8} = 0. \end{aligned}$$
   Both conditions of the question hold: the derivative is right (the constant $-\frac{129}{8}$ differentiates to 0) and the value at 2 is 0.
   $\frac{d}{dx}x^m = m x^{m-1}$⁠, so $\frac{d}{dx}x^{-3} = -3x^{-4}$⁠. Property I, Section 7.2.1.

5. Examine every option

   A correct option must pass two tests: its derivative must be $4x^3 - \frac{3}{x^4}$⁠, and its value at $x = 2$ must be 0.

   Option	Derivative	Right derivative?	Value at 2
   (A)	$4x^3 - \frac{3}{x^4}$	Yes	$0$
   (B)	$3x^2 - \frac{4}{x^5}$	No	$\frac{387}{16} = 24.1875$
   (C)	$4x^3 - \frac{3}{x^4}$	Yes	$\frac{129}{4} = 32.25$
   (D)	$3x^2 - \frac{4}{x^5}$	No	$-\frac{129}{16} = -8.0625$

   (B) and (D) keep the exponents 3 and $-4$ of $f'$ ($x^3$ and $x^{-4}$⁠) instead of raising them to 4 and $-3$⁠; their derivatives are wrong, and neither is 0 at 2 anyway. (C) has the right derivative but the wrong sign on the constant: it gives $f(2) = \frac{129}{8} + \frac{129}{8} \ne 0$⁠. Only (A) passes both tests.
   Values: (B) $8 + \frac{1}{16} + \frac{129}{8} = \frac{387}{16}$⁠; (D) $8 + \frac{1}{16} - \frac{129}{8} = -\frac{129}{16}$⁠.

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