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

The normal of h(x)=3e^x-7x,(0,3) is h(x)= 1/4 x+3

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

normal of

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

h (x) = \frac{1}{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

Compute the slope of the perpendicular line:

m_{p} = \frac{1}{4}

Find the line with slope m=

\frac{1}{4}

and passing through

(0, 3) : h (x) = \frac{1}{4} x + 3

h (x) = \frac{1}{4} x + 3

\frac{1}{2 s + 5}

x \to + 0 lim (cos^{2} (x))

x \to - \infty lim (- 3)

\frac{\partial}{\partial x} (\frac{4}{3} π (y^{3} - x^{3}))

x^{6 c o s (x)}

What is the normal of h(x)=3e^x-7x,(0,3) ?
The normal of h(x)=3e^x-7x,(0,3) is h(x)= 1/4 x+3