What is the normal of y=x^4+3e^x,(0,3) ?

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

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normal of y=x^4+3e^x,(0,3)

normal of

y = x^{4} + 3 e^{x}, (0, 3)

y = - \frac{1}{3} x + 3

Solution steps

Find the slope of

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

Compute the slope of

y = x^{4} + 3 e^{x}

(0, 3) : m = 3

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{3}

Find the line with slope m=

- \frac{1}{3}

and passing through

(0, 3) : y = - \frac{1}{3} x + 3

y = - \frac{1}{3} x + 3

\frac{x ^{3} - 3}{x ^{2} + 2}

\int_{1}^{5} x d x

\int x^{2} (ln (x))^{2} d x

\frac{\partial}{\partial x} (e^{- (x - 3)^{2}})

\frac{d}{d y} (\frac{1}{8} y^{4} + \frac{1}{4 y ^{2}})

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