What is the tangent of f(x)=(4x-2)^{1/5},\at x=0 ?

The tangent of f(x)=(4x-2)^{1/5},\at x=0 is y=(2*2^{1/5})/(5(-1)^{4/5)}x-2^{1/5}

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tangent of f(x)=(4x-2)^{1/5},\at x=0

tangent of

f (x) = (4 x - 2)^{\frac{1}{5}}, at x = 0

y = \frac{2 \cdot 2 ^{\frac{1}{5}}}{5 ( - 1 ) ^{\frac{4}{5}}} x - 2^{\frac{1}{5}}

Solution steps

Find the tangent point:

(0, - 2^{\frac{1}{5}})

Find the slope of

f (x) = (4 x - 2)^{\frac{1}{5}} : \frac{df ( x )}{d x} = \frac{4}{5 ( 4 x - 2 ) ^{\frac{4}{5}}}

Compute the slope of

f (x) = (4 x - 2)^{\frac{1}{5}}

x = 0 : m = \frac{2 \cdot 2 ^{\frac{1}{5}}}{5 ( - 1 ) ^{\frac{4}{5}}}

Find the line with slope m=

\frac{2 \cdot 2 ^{\frac{1}{5}}}{5 ( - 1 ) ^{\frac{4}{5}}}

and passing through

(0, - 2^{\frac{1}{5}}) : y = \frac{2 \cdot 2 ^{\frac{1}{5}}}{5 ( - 1 ) ^{\frac{4}{5}}} x - 2^{\frac{1}{5}}

y = \frac{2 \cdot 2 ^{\frac{1}{5}}}{5 ( - 1 ) ^{\frac{4}{5}}} x - 2^{\frac{1}{5}}

\int 2 x^{3} + x d x

f (x) = 2 x^{2} + 5 x - 3, at x = 1

\int \frac{sin ( 2 x )}{41 + cos ^{2} ( x )} d x

\int_{1}^{3} x ln (x) d x

\frac{\partial}{\partial y} (π x x^{2} + y^{2})

What is the tangent of f(x)=(4x-2)^{1/5},\at x=0 ?
The tangent of f(x)=(4x-2)^{1/5},\at x=0 is y=(2*2^{1/5})/(5(-1)^{4/5)}x-2^{1/5}