What is the tangent of f(x)=x^4-10x^2+9,\at x=1 ?

The tangent of f(x)=x^4-10x^2+9,\at x=1 is y=-16x+16

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tangent of f(x)=x^4-10x^2+9,\at x=1

tangent of

f (x) = x^{4} - 10 x^{2} + 9, at x = 1

y = - 16 x + 16

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = x^{4} - 10 x^{2} + 9 : \frac{df ( x )}{d x} = 4 x^{3} - 20 x

Compute the slope of

f (x) = x^{4} - 10 x^{2} + 9

x = 1 : m = - 16

Find the line with slope m=

- 16

and passing through

(1, 0) : y = - 16 x + 16

y = - 16 x + 16

h (g (x) k (x))

\frac{2}{3} (1 + x)^{\frac{3}{2}}

f (x) = 3 e^{x}, at x = 0

y^{^{'}} = - 5 y + 5 t^{2} + 2 t, y (0) = \frac{1}{3}

\int \frac{3}{4 - x ^{2}} d x

What is the tangent of f(x)=x^4-10x^2+9,\at x=1 ?
The tangent of f(x)=x^4-10x^2+9,\at x=1 is y=-16x+16