{"question":{"question":{"content":"$f(x)=3x^4-16x^3+24x^2+48$. **On which intervals is the graph of $f$ concave down?** [[☃ radio 1]] ","images":{},"widgets":{"radio 1":{"type":"radio","alignment":"default","static":false,"graded":true,"options":{"choices":[{"content":"$\dfrac{2}{3}2$"},{"isNoneOfTheAbove":false,"content":"$x<-2$ and $x>-\dfrac{2}{3}$"},{"isNoneOfTheAbove":false,"content":"$x>0$ only"}],"randomize":true,"multipleSelect":false,"displayCount":null,"hasNoneOfTheAbove":false,"onePerLine":true,"deselectEnabled":false},"version":{"major":1,"minor":0}}}},"answerArea":{"calculator":false,"chi2Table":false,"periodicTable":false,"tTable":false,"zTable":false},"itemDataVersion":{"major":0,"minor":1},"hints":[{"replace":false,"content":"We can analyze the intervals where $f$ is concave up/down by looking for the intervals where its *second* derivative $f''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $f'$, we are analyzing $f''$.","images":{},"widgets":{}},{"replace":false,"content":"The second derivative of $f$ is $f''(x)=12(3x-2)(x-2)$. [[☃ explanation 1]] - $f''(x)=0$ for $x=\dfrac{2}{3},2$. - Since $f''$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=\dfrac{2}{3}$ and $x=2$.","images":{},"widgets":{"explanation 1":{"type":"explanation","alignment":"default","static":false,"graded":true,"options":{"static":false,"showPrompt":"Show entire calculation.","hidePrompt":"Got it, thanks!","explanation":"First, we differentiate $f$ to find $f'$: $\begin{align} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}(3x^4-16x^3+24x^2+48) \\\\ &=3\dfrac{d}{dx}(x^4)-16\dfrac{d}{dx}(x^3)+24\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(48) \\\\ &=3(4x^3)-16(3x^2)+24(2x)+(0) \\\\ &=12x^3-48x^2+48x \end{align}$ Now we can differentiate $f'$ to find $f''$: $\begin{align} &\phantom{=}f''(x) \\\\ &=\dfrac{d}{dx}(12x^3-48x^2+48x) \\\\ &=12\dfrac{d}{dx}(x^3)-48\dfrac{d}{dx}(x^2)+48\dfrac{d}{dx}(x) \\\\ &=12(3x^2)-48(2x)+48(1) \\\\ &=36x^2-96x+48 \\\\ &=12(3x^2-8x+4) \\\\ &=12(3x-2)(x-2) \end{align}$","widgets":{}},"version":{"major":0,"minor":0}}}},{"replace":false,"content":"Our critical points divide the number line into three intervals: ![number line](web+graphie://ka-perseus-graphie.s3.amazonaws.com/9dc3b2cfe20d490cb10b3979728722f29c549aa2) Let's evaluate $f''$ at each interval to see if it's positive or negative on that interval. Interval | $x$-value | $f''(x)$ | Verdict :-: | :-: | :-: | :-: $x<\dfrac{2}{3}$ | $x=0$ | $f''(0)=48>0$ | $f$ is concave up $\cup$ $\dfrac{2}{3}2$ | $x=3$ | $f''(3)=84>0$ | $f$ is concave up $\cup$","images":{"web+graphie://ka-perseus-graphie.s3.amazonaws.com/9dc3b2cfe20d490cb10b3979728722f29c549aa2":{"width":460,"height":80}},"widgets":{}},{"replace":false,"content":"In conclusion, the graph of $f$ is concave down over the interval $\dfrac{2}{3}

Analyze concavity (algebraic)

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