Is #f(x)=1-x-e^(-3x)/x# concave or convex at #x=4#?

Answer 1

Let's take some derivatives!

For #f(x) = 1 - x - e^(-3x)/x#, we have #f'(x) = - 1 - (-3xe^(-3x)-e^(-3x))/x^2# This simplifies (sort of) to #f'(x) = - 1 + e^(-3x)(3x+1)/x^2# Therefore #f''(x) = e^(-3x)(-3x-2)/x^3-3e^(-3x)(3x+1)/x^2# # = e^(-3x)((-3x-2)/x^3-3(3x+1)/x^2)# # = e^(-3x)((-3x-2)/x^3+(-9x-3)/x^2)# # = e^(-3x)((-3x-2)/x^3+(-9x^2-3x)/x^3)# # = e^(-3x)((-9x^2-6x-2)/x^3)#

Now let x = 4.

#f''(4) = e^(-12)((-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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Answer 2

Charles Arocha

To determine whether $f(x) = 1 - x - \frac{e^{-3x}}{x}$ is concave or convex at $x = 4$ , we need to examine the second derivative of the function at that point. If the second derivative is positive, the function is convex; if it's negative, the function is concave.

First, find the first derivative of $f(x)$ : $f'(x) = -1 + \frac{3e^{-3x}}{x^2} + \frac{e^{-3x}(3x+1)}{x^2}$

Then, find the second derivative: $f''(x) = \frac{9e^{-3x}}{x^3} - \frac{6e^{-3x}}{x^2} - \frac{6e^{-3x}(3x+1)}{x^3} + \frac{2e^{-3x}(3x+1)}{x^2}$

Now, evaluate $f''(4)$ to determine the concavity or convexity at $x = 4$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.