What is the tangent of f(x)=7e^x+5x,\at x=0 ?

The tangent of f(x)=7e^x+5x,\at x=0 is y=12x+7

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tangent of f(x)=7e^x+5x,\at x=0

tangent of

f (x) = 7 e^{x} + 5 x, at x = 0

y = 12 x + 7

Solution steps

Find the tangent point:

(0, 7)

Find the slope of

f (x) = 7 e^{x} + 5 x : \frac{df ( x )}{d x} = 7 e^{x} + 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 = 12

Find the line with slope m=

12

and passing through

(0, 7) : y = 12 x + 7

y = 12 x + 7

\int e^{t} \cdot sin (t) d t

n = 1 \sum \infty 3^{- n}

\int \frac{1}{6 + 2 x} d x

\frac{\partial}{\partial x} (cos (x y) sin (2 y))

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

What is the tangent of f(x)=7e^x+5x,\at x=0 ?
The tangent of f(x)=7e^x+5x,\at x=0 is y=12x+7