What is the tangent of f(x)=x^4+9x^2-x,\at x=1 ?

The tangent of f(x)=x^4+9x^2-x,\at x=1 is y=21x-12

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tangent of f(x)=x^4+9x^2-x,\at x=1

tangent of

f (x) = x^{4} + 9 x^{2} - x, at x = 1

y = 21 x - 12

Solution steps

Find the tangent point:

(1, 9)

Find the slope of

f (x) = x^{4} + 9 x^{2} - x : \frac{df ( x )}{d x} = 4 x^{3} + 18 x - 1

Compute the slope of

f (x) = x^{4} + 9 x^{2} - x

x = 1 : m = 21

Find the line with slope m=

21

and passing through

(1, 9) : y = 21 x - 12

y = 21 x - 12

n = 3 \sum \infty (\frac{4}{7})^{n}

\int \frac{1}{cos ^{2} ( 4 x )} d x

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

\int \frac{x ^{2} + 2 x + 5}{( x + 1 ) ( x ^{2} + 1 )} d x

\int_{0}^{1} \frac{12}{x ^{2} - 4} d x

What is the tangent of f(x)=x^4+9x^2-x,\at x=1 ?
The tangent of f(x)=x^4+9x^2-x,\at x=1 is y=21x-12