Find out the answer to "Is f(x)=1−x−e−3xx concave or convex at x=4?"

Annie Rice

Answered question

2022-04-15

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

concave or convex at

x = 4

Answer & Explanation

Austin Sherman

Beginner2022-04-16Added 12 answers

Step 1
For

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

, we have

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

This simplifies (sort of) to

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

Therefore,

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

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

= e^{- 3 x} (\frac{- 3 x - 2}{x^{3}} + \frac{- 9 x - 3}{x^{2}})

= e^{- 3 x} (\frac{- 3 x - 2}{x^{3}} + \frac{- 9 x^{2} - 3 x}{x^{3}})

= e^{- 3 x} (\frac{- 9 x^{2} - 6 x - 2}{x^{3}})

Step 2
Now let

x = 4

f 4) = e^{- 12} (\frac{- 9 {(16)}^{2} - 6 (4) - 2}{4^{3}})

Observe that the exponential is always positive. The numerator of the fraction is negative for all positive values of x. The denominator is positive for positive values of x.
Therefore,

f 4) < 0

Draw your conclusion about concavity.

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