What is the tangent of f(x)=12e^{-0.6x},\at x=9 ?

The tangent of f(x)=12e^{-0.6x},\at x=9 is y=-0.03251…x+(12)/(e^{5.4)}+0.29267…

tangent of f(x)=12e^{-0.6x},\at x=9

tangent of

f (x) = 12 e^{- 0.6 x}, at x = 9

y = - 0.03251 \dots x + \frac{12}{e ^{5.4}} + 0.29267 \dots

Solution steps

Find the tangent point:

(9, \frac{12}{e ^{5.4}})

Find the slope of

f (x) = 12 e^{- 0.6 x} : \frac{df ( x )}{d x} = - 7.2 e^{- 0.6 x}

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 = - 0.03251 \dots

Find the line with slope m=

- 0.03251 \dots

and passing through

(9, \frac{12}{e ^{5.4}}) : y = - 0.03251 \dots x + \frac{12}{e ^{5.4}} + 0.29267 \dots

y = - 0.03251 \dots x + \frac{12}{e ^{5.4}} + 0.29267 \dots

in v erse l a pl a ce \frac{e ^{- 5 s}}{s + 2}

\int \frac{1}{64 e ^{- 4 x} + e ^{4 x}} d x

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

\int \frac{4}{( x ln ( x ) )} d x

s l o p e y = x^{2} - 2 x - 2, (3, 1)

What is the tangent of f(x)=12e^{-0.6x},\at x=9 ?
The tangent of f(x)=12e^{-0.6x},\at x=9 is y=-0.03251…x+(12)/(e^{5.4)}+0.29267…