{"question":{"question":{"content":"The graph of the function $~f(x)~$ is shown. **Which statement is *false* ?** ![](https://ka-perseus-graphie.s3.amazonaws.com/07f270c59fad2539152de8d1dfe021da8e4772d5.png) [[☃ radio 1]]","images":{"https://ka-perseus-graphie.s3.amazonaws.com/07f270c59fad2539152de8d1dfe021da8e4772d5.png":{"width":425,"height":425}},"widgets":{"radio 1":{"type":"radio","graded":true,"options":{"choices":[{"content":"$f\,^{\prime\prime}(-4.5)+ f\,^{\prime\prime}(3)<0$","correct":false,"clue":"The function is concave down at $x=-4.5$ and at $x=3$ which makes $f^{\prime\prime}(-5)<0$ and $f^{\prime\prime}(3)<0$. Since the sum of two negative numbers is negative, this is a **true** statement."},{"content":"$f\,^\prime(-1)+ f\,^{\prime\prime}(-1)<0$","correct":false,"clue":"The function is decreasing and concave down at $x=-1$ which makes $f\,^\prime(-1)<0$ and $f^{\prime\prime}(-1)<0$. Since the sum of two negative numbers is negative, this is a **true** statement."},{"content":"$f\,^\prime(1)+ f\,^{\prime\prime}(1)>0$","correct":false,"clue":"The function is increasing and concave up at $x=1$ which makes $f\,^\prime(1)>0$ and $f^{\prime\prime}(1)>0$. Since the sum of two positive numbers is positive, this is a **true** statement."},{"content":"$f\,^\prime(5)\cdot f\,^{\prime\prime}(5)>0$","correct":true,"clue":"The function is decreasing and concave up at $x=5$ which makes $f\,^\prime(5)<0$ and $f^{\prime\prime}(5)>0$. Since the product of a negative number with a positive number is negative, this is a **false** statement."},{"content":"$f\,^{\prime\prime}(-2)\cdot f\,^{\prime\prime}(2)\approx 0$","correct":false,"clue":"Near $x=-2$ and near $x=2$ the slope along the curve displays little change so $f^{\prime\prime}(-2)\approx 0$ and $f^{\prime\prime}(2)\approx 0$. Since the product of two numbers that are close to zero in value is closer still to zero in value, this is a **true** statement."}],"randomize":false,"multipleSelect":false,"displayCount":null,"hasNoneOfTheAbove":false,"onePerLine":true,"deselectEnabled":false},"version":{"major":1,"minor":0}}}},"answerArea":{"type":"multiple","options":{"content":"","images":{},"widgets":{}},"calculator":false},"itemDataVersion":{"major":0,"minor":1},"hints":[{"content":"This problem requires that we use the following ideas. I. $~~~~~~$If $~f~$ is increasing at $~x=a\,$, then $f\,^\prime(a)>0\,$. II. $~~~~~$If $~f~$ is decreasing at $~x=a\,$, then $~f\,^\prime(a)<0\,$. III. $~~~~$If $~f~$ is concave up at $~x=a\,$, then $~f\,^{\prime\prime}(a)>0\,$. IV. $~~~$If $~f~$ is concave down at $~x=a\,$, then $~f\,^{\prime\prime}(a)<0\,$. V. $~~~$On sections of graphs where the slopes look nearly constant (i.e. the graph looks linear), then the second derivative will be $~0~$ or close to $~0\,$.","images":{},"widgets":{}},{"content":"The first choice. The graph of $~f~$ is concave down at $~x=-4.5~$ and again at $~x=3\,$. Hence the given expression is the sum of two negative numbers, which is indeed negative. So it follows that $~f\,^{\prime\prime}(-4.5)+ f\,^{\prime\prime}(3)<0~$ is **true**.","images":{},"widgets":{}},{"content":"The second choice. The graph of $~f~$ is both decreasing and concave down at $~x=-1\,$. Hence the given expression is the sum of two negative numbers, which is indeed negative. $~f\,^\prime(-1)+ f\,^{\prime\prime}(-1)<0~$ is **true**.","images":{},"widgets":{}},{"content":"The third choice. The graph of $~f~$ is both increasing and concave up at $~x=1\,$. Hence $~f\,^\prime(1)~$ and $~f\,^{\prime\prime}(1)~$ are both positive. The expression is the sum of two positive numbers which is, of course, positive. Therefore $~f\,^\prime(1)+ f\,^{\prime\prime}(1)>0~$ is **true**.","images":{},"widgets":{}},{"content":"The fourth choice. The graph of $~f~$ is decreasing and concave up at $~x=5\,$. Hence the given expression is the product of a negative number and a positive number, which is negative. So it follows that $~f\,^\prime(5)\cdot f^{\prime\prime}(5)>0~$ is **false**.","images":{},"widgets":{}},{"content":"The fifth choice. The graph of $~f~$ at $~x=-2~$ and again at $~x=2~$ is quite linear making the second derivative values close to zero at each of those $~x$-values. Since each of those numbers is close to zero, their product is **very** close to zero. So $~f\,^{\prime\prime}(-2)\cdot f\,^{\prime\prime}(2)\approx 0~$ is **true**.","images":{},"widgets":{}}]},"problemId":40239,"exerciseId":3299,"exerciseSessionId":null,"hintsEnabled":false,"readOnly":false,"submitterId":null,"checkAnswerButtonText":"Check answer","correctButtonText":"🌟 Yes! You got it!","tryAgainButtonText":"Try again","problemAiConversationId":null}

Concavity & inflection points challenge

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