What is the tangent of f(x)=15e^x+11x,\at x=0 ?

The tangent of f(x)=15e^x+11x,\at x=0 is y=26x+15

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tangent of f(x)=15e^x+11x,\at x=0

tangent of

f (x) = 15 e^{x} + 11 x, at x = 0

y = 26 x + 15

Solution steps

Find the tangent point:

(0, 15)

Find the slope of

f (x) = 15 e^{x} + 11 x : \frac{df ( x )}{d x} = 15 e^{x} + 11

Compute the slope of

f (x) = 15 e^{x} + 11 x

x = 0 : m = 26

Find the line with slope m=

26

and passing through

(0, 15) : y = 26 x + 15

y = 26 x + 15

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What is the tangent of f(x)=15e^x+11x,\at x=0 ?
The tangent of f(x)=15e^x+11x,\at x=0 is y=26x+15