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

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

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

tangent of

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

Compute the slope of

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

x = 2 : m = 1

Find the line with slope m=

1

and passing through

(2, 2) : y = x

y = x

x \cdot y^{^{'}} = 2 y + e^{x} \cdot x^{3}

x \to - \frac{1}{2} lim (\frac{7}{2 x + 1})

- x^{2} + 2 x, 3 x^{3} - x^{2} - 10 x

5 sin (- 3 x + \frac{π}{4})

x \to 3 lim (5^{2} - 9 x + 7)

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