What is the tangent of f(x)=3x^4-8x^2,(-1,5) ?

The tangent of f(x)=3x^4-8x^2,(-1,5) is y=-4.38103…x+0.61896… \lor y=-5.01374…x-0.01374…

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tangent of f(x)=3x^4-8x^2,(-1,5)

tangent of

f (x) = 3 x^{4} - 8 x^{2}, (- 1, 5)

y = - 4.38103 \dots x + 0.61896 \dots or y = - 5.01374 \dots x - 0.01374 \dots

Solution steps

Slope of the tangent line at

(- 1, 5) : m = \frac{3 x ^{4} - 8 x ^{2} - 5}{x + 1}

Find the slope of

f (x) = 3 x^{4} - 8 x^{2} : m = 12 x^{3} - 16 x

Solve

12 x^{3} - 16 x = \frac{3 x ^{4} - 8 x ^{2} - 5}{x + 1} : x \approx 0.29260 \dots, x \approx 0.94371 \dots

Find the tangent line at

x = 0.29260 \dots : y = - 4.38103 \dots x + 0.61896 \dots

Find the tangent line at

x = 0.94371 \dots : y = - 5.01374 \dots x - 0.01374 \dots

y = - 4.38103 \dots x + 0.61896 \dots or y = - 5.01374 \dots x - 0.01374 \dots

\frac{\partial}{\partial x} (- 3 x y ln (x y))

\frac{\partial}{\partial x} (\frac{x y ^{2}}{z ^{3}})

y^{^{''}} = - y

\int x - 7 d x

x \to + 0 lim (\frac{e ^{x^{2}} - 1}{x})

What is the tangent of f(x)=3x^4-8x^2,(-1,5) ?
The tangent of f(x)=3x^4-8x^2,(-1,5) is y=-4.38103…x+0.61896… \lor y=-5.01374…x-0.01374…