Question

1. if ( f(x) = -x^3 + x + \frac{1}{x} ), then ( f(-1) = )

no calculator
(a) 3 (b) 1 (c) -1 (d) -3 (e) -5

Explanation:

Step1: Differentiate each term

For \( f(x) = -x^3 + x + \frac{1}{x} \), we can rewrite \( \frac{1}{x} \) as \( x^{-1} \).
Using the power rule \( \frac{d}{dx}(x^n) = nx^{n - 1} \):

• The derivative of \( -x^3 \) is \( -3x^2 \) (since \( n = 3 \), so \( 3x^{3 - 1}=3x^2 \), and with the negative sign, it's \( -3x^2 \)).
• The derivative of \( x \) (where \( n = 1 \)) is \( 1\times x^{1 - 1}=1 \).
• The derivative of \( x^{-1} \) (where \( n=- 1\)) is \( -1\times x^{-1 - 1}=-x^{-2}=-\frac{1}{x^2} \).

So, \( f'(x)=-3x^2 + 1-\frac{1}{x^2} \).

Step2: Substitute \( x = - 1 \) into \( f'(x) \)

Substitute \( x=-1 \) into \( f'(x) \):
\( f'(-1)=-3\times(-1)^2 + 1-\frac{1}{(-1)^2} \)
First, calculate \( (-1)^2 = 1 \):
\( f'(-1)=-3\times1 + 1-\frac{1}{1} \)
Then, simplify each term:
\( -3\times1=-3 \), \( \frac{1}{1} = 1 \)
So, \( f'(-1)=-3 + 1-1 \)
\( f'(-1)=-3 \)

Answer:

D. -3