What is the equation of the line tangent to f(x)=1ex−3x at x=0?

Explanation:At x=0, the function has y-valuef(0)=1e0−3(0)=11=1We now rewrite the function asf(x)=(ex−3x)−12We can find this derivative using the chain rule.f′(x)=−12u−32⋅(ex−3)=−ex−32(ex−3x)32The slope of the tangent is hencef′(0)=−e0−32(e0−3(0))32f′(0)=−−22f′(0)=22f'(0)=1Now write the equation of the tangent.y−y1=m(x−x1)y-1=1(x-0)y=x+1

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gabolzm6d

Answered question

2022-04-07

What is the equation of the line tangent to

f (x) = \frac{1}{\sqrt{e^{x} - 3 x}}

at x=0?

Answer & Explanation

Videoad3u

Beginner2022-04-08Added 15 answers

Explanation:
At x=0, the function has y-value

f (0) = \frac{1}{\sqrt{e^{0} - 3 (0)}} = \frac{1}{\sqrt{1}} = 1

We now rewrite the function as

f (x) = {(e^{x} - 3 x)}^{- \frac{1}{2}}

We can find this derivative using the chain rule.

f^{'} (x) = - \frac{1}{2} u^{- \frac{3}{2}} \cdot (e^{x} - 3) = - \frac{e^{x} - 3}{2 {(e^{x} - 3 x)}^{\frac{3}{2}}}

The slope of the tangent is hence

f^{'} (0) = - \frac{e^{0} - 3}{2 {(e^{0} - 3 (0))}^{\frac{3}{2}}}

f^{'} (0) = - \frac{- 2}{2}

f^{'} (0) = \frac{2}{2}

f'(0)=1
Now write the equation of the tangent.

y - y_{1} = m (x - x_{1})

y-1=1(x-0)
y=x+1

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