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Let $f(x) = \begin{cases} x^2+4x, & x<1; \\ x^3-6x+10, & x\geq1. \end{cases}$

Find the critical points.

I have solved. The function is not differentiable at $x=1$. So $x=1$ is a critical point. Again by setting $f'(x)=0$, we get $-2$, $-\sqrt{2}$, $1$, $\sqrt{2}$ are the critical points. But when I enter the number in the machine, it says its wrong. Kindly help.

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Your answers are correct... almost. The one you have wrong is $-\sqrt{2}$. You got that from the part of $f(x)$ that was only defined when $x>1$. You can't have a critical point of a graph that isn't defined, that is, doesn't exist. Here, check out the graph of your function. https://www.desmos.com/calculator/ju77wsjwbk

I mean, you got it from the $f(x) = x^3 + 6x + 10$ but for that part $x$ has to be greater than 1.

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For $x<1$ we have $x_{min}=-2$ and $(-2,-4)$ it's a minimum point.

$f$ increases on $[-2,1]$ and decreases on $[1,\sqrt2]$, which says that $x_{max}=1$ and $(1,5)$ is a maximum point.

For $x\geq1$ we have $f'(x)=3x^2-6=3(x-\sqrt2)(x+\sqrt2),$ which gives $x_{min}=\sqrt2$

and $(\sqrt2,10-4\sqrt2)$ is a minimum point.

$x=-\sqrt2$ it's nothing of course.

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Don't forget that this is a piecewise function!

You corectly observed that one of the critical points in $x=1$ because $f'(x)$ doesn't exist at this point. Then you proceed to finding the "usual" critical points, by setting up $f'(x)=0$, on each of the two pieces of the function. Remember that whatever you find for a piece must belong to that piece!

  • For the piece $f(x)=x^2+4x$ on $x<1$, only the points that actually belong to the interval $x<1$ are meaningful.

  • For the piece $f(x)=x^3-6x+10$ on $x\ge1$, only the points that actually belong to the interval $x\ge1$ are meaningful.

Your answer is incorrect because one of the points that you found can't be included since it doesn't lie within the respective interval.

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