What is the tangent of f(x)=18e^x+10x,\at x=0 ?

The tangent of f(x)=18e^x+10x,\at x=0 is y=28x+18

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tangent of f(x)=18e^x+10x,\at x=0

tangent of

f (x) = 18 e^{x} + 10 x, at x = 0

y = 28 x + 18

Solution steps

Find the tangent point:

(0, 18)

Find the slope of

f (x) = 18 e^{x} + 10 x : \frac{df ( x )}{d x} = 18 e^{x} + 10

Compute the slope of

f (x) = 18 e^{x} + 10 x

x = 0 : m = 28

Find the line with slope m=

28

and passing through

(0, 18) : y = 28 x + 18

y = 28 x + 18

\int \frac{6 x}{x ^{3} - 8} d x

\int sec^{2} (π x) d x

\int_{- 1}^{1} t (1 - t)^{2} d t

x \to 0 lim (x^{3} ln (x^{2} + 1))

n = 1 \sum \infty \frac{( - 3 ) ^{n - 1}}{8 ^{n}}

What is the tangent of f(x)=18e^x+10x,\at x=0 ?
The tangent of f(x)=18e^x+10x,\at x=0 is y=28x+18