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

The tangent of f(x)=10-e^x,\at x=0 is y=-x+9

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

tangent of

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

y = - x + 9

Solution steps

Find the tangent point:

(0, 9)

Find the slope of

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

Compute the slope of

f (x) = 10 - e^{x}

x = 0 : m = - 1

Find the line with slope m=

- 1

and passing through

(0, 9) : y = - x + 9

y = - x + 9

(\frac{1}{t + 1})^{^{'}}

\frac{\partial}{\partial v} (u - v)

\int - 3 x^{2} - 12 x + 15 d x

\int_{0}^{1} \frac{3}{x ^{5}} d x

\int (e^{x} + sin (x) + x) d x

What is the tangent of f(x)=10-e^x,\at x=0 ?
The tangent of f(x)=10-e^x,\at x=0 is y=-x+9