What is the tangent of f(x)=x^4-5x^2+2x-4,(1,0) ?

The tangent of f(x)=x^4-5x^2+2x-4,(1,0) is y=3.83118…x-3.83118… \lor y=5.31735…x-5.31735…

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tangent of f(x)=x^4-5x^2+2x-4,(1,0)

tangent of

f (x) = x^{4} - 5 x^{2} + 2 x - 4, (1, 0)

y = 3.83118 \dots x - 3.83118 \dots or y = 5.31735 \dots x - 5.31735 \dots

Solution steps

Slope of the tangent line at

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

Find the slope of

f (x) = x^{4} - 5 x^{2} + 2 x - 4 : m = 4 x^{3} - 10 x + 2

Solve

4 x^{3} - 10 x + 2 = \frac{x ^{4} - 5 x ^{2} + 2 x - 4}{x - 1} : x \approx - 0.18567 \dots, x \approx - 1.37768 \dots

Find the tangent line at

x = - 0.18567 \dots : y = 3.83118 \dots x - 3.83118 \dots

Find the tangent line at

x = - 1.37768 \dots : y = 5.31735 \dots x - 5.31735 \dots

y = 3.83118 \dots x - 3.83118 \dots or y = 5.31735 \dots x - 5.31735 \dots

n = 1 \sum \infty \frac{n}{n ^{3} + 5}

\frac{d}{d x} (x 4 - x)

\int \frac{7}{9 + 16 x ^{2}} d x

(\frac{sin ( x )}{1 + cos ( x )})^{2}

\int x 3 x d x

What is the tangent of f(x)=x^4-5x^2+2x-4,(1,0) ?
The tangent of f(x)=x^4-5x^2+2x-4,(1,0) is y=3.83118…x-3.83118… \lor y=5.31735…x-5.31735…