What is the tangent of f(x)=14-e^x ?

The tangent of f(x)=14-e^x is y=-e^{a_{0}}x+14-e^{a_{0}}+a_{0}e^{a_{0}}

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tangent of f(x)=14-e^x

tangent of

f (x) = 14 - e^{x}

y = - e^{a_{0}} x + 14 - e^{a_{0}} + a_{0} e^{a_{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 14 - e^{a_{0}})

Find the slope of

f (x) = 14 - e^{x} : \frac{df ( x )}{d x} = - e^{x}

Compute the slope of

f (x) = 14 - e^{x}

x = a_{0} : m = - e^{a_{0}}

Find the line with slope m=

- e^{a_{0}}

and passing through

(a_{0}, 14 - e^{a_{0}}) : y = - e^{a_{0}} x + 14 - e^{a_{0}} + a_{0} e^{a_{0}}

y = - e^{a_{0}} x + 14 - e^{a_{0}} + a_{0} e^{a_{0}}

f (x) = 2 x + 1

(2 x - 7)^{3}

x \to 0 lim (\frac{sec ^{2} ( 3 x )}{2})

f (x) = x^{3} + x^{2}

\frac{d}{d x} (\frac{1}{3} cos^{3} (x))

What is the tangent of f(x)=14-e^x ?
The tangent of f(x)=14-e^x is y=-e^{a_{0}}x+14-e^{a_{0}}+a_{0}e^{a_{0}}