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Rom4ik [11]
3 years ago
5

PLS HELP THIS IS HARD ANYONE PLS

Mathematics
1 answer:
aalyn [17]3 years ago
4 0

Answer:

a or c

Step-by-step explanation:

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c.

Step-by-step explanation:

1.25x5=6.25

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Finding an Equation of a tangent Line in Exercise, find an equation of the tangent line to the graph of the function at the give
frutty [35]

Answer:

y=\dfrac{3x}{e}+\dfrac{4}{e}

this is the equation of the tangent at point (-1,1/e)

Step-by-step explanation:

to find the tangent line we need to find the derivative of the function g(x).

g(x) =e^{x^3}

  • we know that \frac{d}{dx}(e^{f(x)})=e^{f(x)}f'(x)

g'(x) =e^{x^{3}}(3 x^{2})

g'(x) =3 x^{2} e^{x^{3}}

this the equation of the slope of the curve at any point x and it also the slope of the tangent at any point x. hence, g'(x) can be denoted as 'm'

to find the slope at (-1,1/e) we'll use the x-coordinate of the point i.e. x = -1

m =3 (-1)^{2} e^{(-1)^{3}}\\m =3e^{-1}\\m=\dfrac{3}{e}

using the equation of line:

(y-y_1)=m(x-x_1)

we'll find the equation of the tangent line.

here (x1,y1) =(-1,1/e), and m = 3/e

(y-\dfrac{1}{e})=\dfrac{3}{e}(x+1)\\y=\dfrac{3x}{e}+\dfrac{3}{e}+\dfrac{1}{e}\\

y=\dfrac{3x}{e}+\dfrac{4}{e}

this is the equation of the tangent at point (-1,1/e)

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