What is the normal of y=x^4+7e^x,(0,7) ?

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

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normal of y=x^4+7e^x,(0,7)

normal of

y = x^{4} + 7 e^{x}, (0, 7)

y = - \frac{1}{7} x + 7

Solution steps

Find the slope of

y = x^{4} + 7 e^{x} : \frac{d y}{d x} = 4 x^{3} + 7 e^{x}

Compute the slope of

y = x^{4} + 7 e^{x}

(0, 7) : m = 7

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{7}

Find the line with slope m=

- \frac{1}{7}

and passing through

(0, 7) : y = - \frac{1}{7} x + 7

y = - \frac{1}{7} x + 7

x \to 2 - lim (H (x))

\frac{d}{d x} (x (2 x + 9)^{\frac{1}{3}})

\int_{0}^{\infty} t e^{- k t} d t

\int \frac{x}{x ^{2} + 2 x + 1} d x

\frac{d}{d x} (2 x^{2} - 4)

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