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

The tangent of f(x)=(2x+4)^{1/5},\at x=1 is y=(2^{1/5})/(5*3^{4/5)}x+6^{1/5}-(2^{1/5})/(5*3^{4/5)}

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

tangent

y = \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}} x + 6^{\frac{1}{5}} - \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}}

Solution steps

Find the tangent point:

(1, 6^{\frac{1}{5}})

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}}

Find the line with slope m=

\frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}}

and passing through

(1, 6^{\frac{1}{5}}) : y = \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}} x + 6^{\frac{1}{5}} - \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}}

y = \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}} x + 6^{\frac{1}{5}} - \frac{2 ^{\frac{1}{5}}}{5 \cdot 3 ^{\frac{4}{5}}}

What is the tangent of f(x)=(2x+4)^{1/5},\at x=1 ?
The tangent of f(x)=(2x+4)^{1/5},\at x=1 is y=(2^{1/5})/(5*3^{4/5)}x+6^{1/5}-(2^{1/5})/(5*3^{4/5)}