What is the tangent of f(x)=3x^5-11x^3,\at x=1 ?

The tangent of f(x)=3x^5-11x^3,\at x=1 is y=-18x+10

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tangent of f(x)=3x^5-11x^3,\at x=1

tangent of

f (x) = 3 x^{5} - 11 x^{3}, at x = 1

y = - 18 x + 10

Solution steps

Find the tangent point:

(1, - 8)

Find the slope of

f (x) = 3 x^{5} - 11 x^{3} : \frac{df ( x )}{d x} = 15 x^{4} - 33 x^{2}

Compute the slope of

f (x) = 3 x^{5} - 11 x^{3}

x = 1 : m = - 18

Find the line with slope m=

- 18

and passing through

(1, - 8) : y = - 18 x + 10

y = - 18 x + 10

\frac{d}{d x} (cot (11 x))

\int (a + b t)^{2} d t

\int_{2}^{5} 4 d x

y^{^{''}} - 12 y^{^{'}} + 35 y = e^{8 x}

y = \frac{7 x}{x}

What is the tangent of f(x)=3x^5-11x^3,\at x=1 ?
The tangent of f(x)=3x^5-11x^3,\at x=1 is y=-18x+10