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

The tangent of f(x)=(4x^2-3x)^2,\at x=1 is y=10x-9

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tangent of f(x)=(4x^2-3x)^2,\at x=1

tangent of

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

y = 10 x - 9

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

f (x) = (4 x^{2} - 3 x)^{2} : \frac{df ( x )}{d x} = 2 (4 x^{2} - 3 x) (8 x - 3)

Compute the slope of

f (x) = (4 x^{2} - 3 x)^{2}

x = 1 : m = 10

Find the line with slope m=

10

and passing through

(1, 1) : y = 10 x - 9

y = 10 x - 9

\int \frac{x ^{2} + 20 x - 4}{x ^{3} - 4 x} d x

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

(12 x^{2} - 20 y^{2}) d x + (- 20 x y + 36 x^{- 1} y^{3}) d y = 0

\int e^{x} d x

\int \frac{x ^{2}}{x + 2} d x

What is the tangent of f(x)=(4x^2-3x)^2,\at x=1 ?
The tangent of f(x)=(4x^2-3x)^2,\at x=1 is y=10x-9