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Question
section 2.7: second derivative and concavity (homew
score: 0/80 answered: 0/8
question 1
if (f(x)=7 + \frac{2}{x}+\frac{6}{x^{2}}), find (f(x)).
find (f(4)).
find (f(x)).
find (f(4)).
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Step1: Rewrite the function
Rewrite $f(x)=7 + \frac{2}{x}+\frac{6}{x^{2}}$ as $f(x)=7 + 2x^{-1}+6x^{-2}$.
Step2: Find the first - derivative
Use the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$.
$f'(x)=\frac{d}{dx}(7)+\frac{d}{dx}(2x^{-1})+\frac{d}{dx}(6x^{-2})=0+2(-1)x^{-2}+6(-2)x^{-3}=- \frac{2}{x^{2}}-\frac{12}{x^{3}}$.
Step3: Find $f'(4)$
Substitute $x = 4$ into $f'(x)$.
$f'(4)=-\frac{2}{4^{2}}-\frac{12}{4^{3}}=-\frac{2}{16}-\frac{12}{64}=-\frac{8 + 12}{64}=-\frac{20}{64}=-\frac{5}{16}$.
Step4: Find the second - derivative
Differentiate $f'(x)=-2x^{-2}-12x^{-3}$ using the power rule.
$f''(x)=\frac{d}{dx}(-2x^{-2})+\frac{d}{dx}(-12x^{-3})=-2(-2)x^{-3}-12(-3)x^{-4}=\frac{4}{x^{3}}+\frac{36}{x^{4}}$.
Step5: Find $f''(4)$
Substitute $x = 4$ into $f''(x)$.
$f''(4)=\frac{4}{4^{3}}+\frac{36}{4^{4}}=\frac{4}{64}+\frac{36}{256}=\frac{16 + 36}{256}=\frac{52}{256}=\frac{13}{64}$.
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$f'(x)=-\frac{2}{x^{2}}-\frac{12}{x^{3}}$
$f'(4)=-\frac{5}{16}$
$f''(x)=\frac{4}{x^{3}}+\frac{36}{x^{4}}$
$f''(4)=\frac{13}{64}$