Question

if $f(x)=6+\frac{4}{x}+\frac{3}{x^2}$, find $f(x)$.
find $f(4)$.
question 6
if $f(x)=3+\frac{4}{x}+\frac{4}{x^2}$, find $f(x)$.
find $f(4)$.
find $f(x)$.
find $f(4)$.

Explanation:

For the first problem ($f(x) = 6 + \frac{4}{x} + \frac{3}{x^2}$):

Step1: Rewrite terms for differentiation

Rewrite the rational terms using negative exponents:
$f(x) = 6 + 4x^{-1} + 3x^{-2}$

Step2: Apply power rule to find $f'(x)$

Power rule: $\frac{d}{dx}[x^n] = nx^{n-1}$
$f'(x) = 0 + 4(-1)x^{-2} + 3(-2)x^{-3} = -\frac{4}{x^2} - \frac{6}{x^3}$

Step3: Substitute $x=4$ into $f'(x)$

Calculate each term separately:
$f'(4) = -\frac{4}{4^2} - \frac{6}{4^3} = -\frac{4}{16} - \frac{6}{64}$

Step4: Simplify the expression

Reduce fractions and combine:
$f'(4) = -\frac{1}{4} - \frac{3}{32} = -\frac{8}{32} - \frac{3}{32} = -\frac{11}{32}$

Step1: Rewrite terms for differentiation

Rewrite rational terms with negative exponents:
$f(x) = 3 + 4x^{-1} + 4x^{-2}$

Step2: Find first derivative $f'(x)$

Apply power rule:
$f'(x) = 0 + 4(-1)x^{-2} + 4(-2)x^{-3} = -\frac{4}{x^2} - \frac{8}{x^3}$

Step3: Calculate $f'(4)$

Substitute $x=4$ into $f'(x)$:
$f'(4) = -\frac{4}{4^2} - \frac{8}{4^3} = -\frac{4}{16} - \frac{8}{64} = -\frac{1}{4} - \frac{1}{8} = -\frac{3}{8}$

Step4: Find second derivative $f''(x)$

Differentiate $f'(x) = -4x^{-2} -8x^{-3}$ using power rule:
$f''(x) = -4(-2)x^{-3} -8(-3)x^{-4} = \frac{8}{x^3} + \frac{24}{x^4}$

Step5: Calculate $f''(4)$

Substitute $x=4$ into $f''(x)$:
$f''(4) = \frac{8}{4^3} + \frac{24}{4^4} = \frac{8}{64} + \frac{24}{256} = \frac{1}{8} + \frac{3}{32} = \frac{7}{32}$

Answer:

$f'(x) = -\frac{4}{x^2} - \frac{6}{x^3}$
$f'(4) = -\frac{11}{32}$

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For the second problem ($f(x) = 3 + \frac{4}{x} + \frac{4}{x^2}$):