Question

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

Answer 1

Let's take some derivatives!

Explanation:

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.