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Consider the function $f\left(x\right)=x^2e^{\left(-x+4\right)}$

The first and second derivatives are:

$f'\left(x\right)=x\left(2-x\right)e^{\left(-x+4\right)}$

$f''\left(x\right)=\left(x^2-4x+2\right)e^{\left(-x+4\right)}$

e) Determine the intervals where $f(x)$ is concave up and concave down

I know that $f''(x) < 0 $ curve is concave down, if $f''(x) > 0$, then the curve is concave down. I also know that if $f''(x) = 0$, then $x$ is the point of inflection, where the concavity changes.

So I thought I compute it: $0=\left(x^2-4x+2\right)e^{\left(-x+4\right)}$

$\:x^2-4x+2=0$

using the quadratic formula:

$x=\frac{-\left(-4\right)-\sqrt{\left(-4\right)^2-4\cdot \:1\cdot \:2}}{2\cdot \:1}:\quad 2-\sqrt{2}$ or $2+\sqrt{2}$

The solution is:

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But why? I don't see it? $2-\sqrt{2}$ to the maximum has a positive gradient right? Then how come is is concave down? What is the general approach to questions that ask something like this?

enter image description here

Probability_Sarah
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  • It's not about the gradient being positive: that's asking for $f'(x) >0$. But the rate of growth of the gradient is what matters. It's not as intuitive, it's a second order thing (literally). – Henno Brandsma Sep 19 '20 at 09:06
  • $f$ being concave up means that the area underneath the graph is convex. Which checks out in our graph, right? – Henno Brandsma Sep 19 '20 at 09:15

2 Answers2

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The sign of $f''(x)$ is determined by the sign of the polynomial (because any $e^A$ is always strictly positive, whatever $A$ is) $x^2 - 2x + 4$ which is a parabola shaped like $x^2$ (what we in Dutch call a "dalparabool" ("valley-parabola", roughly) so positive to the left and right of any zeroes. So $f$ is concave up left of the zero $\frac{4-\sqrt{8}}{2} = 2- \sqrt{2}$, and to the right of the zero $\frac{4+\sqrt{8}}{2} = 2+\sqrt{2}$, just as your solution manual said.

Inbetween the zeroes, the polynomial is $<0$ so $f''(x) < 0$ and so it is concave down. Apply the facts you know. The graph is not needed.

Also for the derivative, whose sign is determined by a polynomial $-x^2 + 2x$ (shaped upside down), we can predict that $f'(x)>0$ is positive (so upward slope) for all values inbetween the zeroes $0$ and $2$ and negative (downward gradient) for $x<0$ and $x>2$. But graphs being concave up/down is determined entirely by the sign of $f''(x)$, so whether the gradient itself is decreasing or increasing (so whether growth slows down or increases).

Henno Brandsma
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If $f’’(x)=(x^2-4x+2)e^{4-x}$ then $f’’(x)>0 \iff x^2-4x+2 >0$. This is true because $e^{4-x}>0$ for any $x\in\mathbb{R}$ and so the sign of $f’’$ is the same as the sign of $x^2-4x+2$.

Can you reach to the conclusions from here?

Henno Brandsma
  • 242,131
  • Thanks Gio! What if the "$e$" doesn't exist, I just substitute numbers in to see if $f''(x)$ is negative or positive or zero? – Probability_Sarah Sep 25 '20 at 05:24
  • If $f’’(x)=x^2-4x+2=(x-2)^2-2$ then to study where $f$ is concave down you would have to solve $f’’(x)<0$,... –  Sep 25 '20 at 10:16