What is the tangent of e^x-2x ?

The tangent of e^x-2x is y=(e^{a_{0}}-2)x-a_{0}e^{a_{0}}+e^{a_{0}}

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tangent of e^x-2x

tangent of

e^{x} - 2 x

y = (e^{a_{0}} - 2) x - a_{0} e^{a_{0}} + e^{a_{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, e^{a_{0}} - 2 a_{0})

Find the slope of

y = e^{x} - 2 x : \frac{d y}{d x} = e^{x} - 2

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 = e^{a_{0}} - 2

Find the line with slope m=

e^{a_{0}} - 2

and passing through

(a_{0}, e^{a_{0}} - 2 a_{0}) : y = (e^{a_{0}} - 2) x - a_{0} e^{a_{0}} + e^{a_{0}}

y = (e^{a_{0}} - 2) x - a_{0} e^{a_{0}} + e^{a_{0}}

\frac{d}{d x} (\frac{2 x ^{2} + 4 x + 6}{x})

\frac{d}{d x} (4 x^{2} - 4 x + 1)

f (x) = arcsin (ln (x))

x \to 1 lim (1 - \frac{1}{3 x - 1})

\int \frac{6}{1 - cos ( 2 x )} d x

What is the tangent of e^x-2x ?
The tangent of e^x-2x is y=(e^{a_{0}}-2)x-a_{0}e^{a_{0}}+e^{a_{0}}