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

The tangent of f(x)=x(x^2-4x+5)^4,\at x=2 is y=x

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tangent of f(x)=x(x^2-4x+5)^4,\at x=2

tangent of

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

y = x

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

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

Compute the slope of

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

x = 2 : m = 1

Find the line with slope m=

1

and passing through

(2, 2) : y = x

y = x

\frac{1}{s ^{2} + 2 s - 3}

\int \frac{e ^{- s t}}{s} d t

\int ln (6 x + 1) d x

\frac{d}{d x} (sin (e^{x} - 1))

4 x^{3} - 4 x^{2} + 3

What is the tangent of f(x)=x(x^2-4x+5)^4,\at x=2 ?
The tangent of f(x)=x(x^2-4x+5)^4,\at x=2 is y=x