What is the tangent of h(x)=3e^x-7x,(0,3) ?

The tangent of h(x)=3e^x-7x,(0,3) is y=-4x+3

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tangent of h(x)=3e^x-7x,(0,3)

tangent of

h (x) = 3 e^{x} - 7 x, (0, 3)

y = - 4 x + 3

Solution steps

Find the slope of

h (x) = 3 e^{x} - 7 x : \frac{d h ( x )}{d x} = 3 e^{x} - 7

Compute the slope of

h (x) = 3 e^{x} - 7 x

(0, 3) : m = - 4

Find the line with slope m=

- 4

and passing through

(0, 3) : y = - 4 x + 3

y = - 4 x + 3

\frac{d}{d x} (e^{t a n (3 x)})

(9, 2), (- 2, 6)

\frac{d}{d x} (\frac{2}{x} - 1)

\frac{( x ^{2} )}{5 + x}

\frac{d y}{d x} + \frac{y}{x} = 2 x^{2} y^{2}

What is the tangent of h(x)=3e^x-7x,(0,3) ?
The tangent of h(x)=3e^x-7x,(0,3) is y=-4x+3