What is the tangent of f(x)=x^3-2/(x^2)+2,(1,3) ?

The tangent of f(x)=x^3-2/(x^2)+2,(1,3) is y=15.59616…x-12.59616… \lor y=1.87099…x+1.12900…

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tangent of f(x)=x^3-2/(x^2)+2,(1,3)

tangent of

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

y = 15.59616 \dots x - 12.59616 \dots or y = 1.87099 \dots x + 1.12900 \dots

Solution steps

Slope of the tangent line at

(1, 3) : m = \frac{x ^{5} - 2 - x ^{2}}{x ^{2} ( x - 1 )}

Find the slope of

f (x) = x^{3} - \frac{2}{x ^{2}} + 2 : m = 3 x^{2} + \frac{4}{x ^{3}}

Solve

3 x^{2} + \frac{4}{x ^{3}} = \frac{x ^{5} - 2 - x ^{2}}{x ^{2} ( x - 1 )} : x \approx 0.65378 \dots, x \approx - 1.18972 \dots

Find the tangent line at

x = 0.65378 \dots : y = 15.59616 \dots x - 12.59616 \dots

Find the tangent line at

x = - 1.18972 \dots : y = 1.87099 \dots x + 1.12900 \dots

y = 15.59616 \dots x - 12.59616 \dots or y = 1.87099 \dots x + 1.12900 \dots

What is the tangent of f(x)=x^3-2/(x^2)+2,(1,3) ?
The tangent of f(x)=x^3-2/(x^2)+2,(1,3) is y=15.59616…x-12.59616… \lor y=1.87099…x+1.12900…