","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\text{{f}'(x)=0}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" The below graph represents an increasing function as the value of $$x$$ increases. $$ \\\\ $$ ","position":1,"answerExplanation":{"@type":"Comment","text":" A function $$f(x)$$ is an increasing function if $${f}'(x) > 0$$ $$ \\\\ $$ The graph of the increasing function goes on increasing as the value of $$x$$ increases.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Increasing and decreasing intervals Quiz 1","text":" Which of the following graphs will represent the graph of decreasing function in the interval $$\\left( 1,\\infty \\right)$$","comment":{"@type":"Comment","text":" A graph goes on increasing in a certain interval then the function is said to be increasing. A graph goes on decreasing in a certain interval then the function is said to be decreasing. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"
","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" Both $$(a)$$ and $$(b)$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" None of the above","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":"
","position":1,"answerExplanation":{"@type":"Comment","text":" An increasing function is a function such that the graph goes on increasing in a respective interval. $$ \\\\ $$ From option A, $$ \\\\ $$ It is clear that the graph goes on increasing in the interval $$\\left( 1,\\infty \\right)$$. $$ \\\\ $$
$$ \\\\ $$ From option B, $$ \\\\ $$ It is clear that the graph goes on decreasing in the interval $$\\left( 1,\\infty \\right)$$
","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Increasing and decreasing intervals Quiz 1","text":" From the given graph, which of the following statements are correct? $$ \\\\ $$
","comment":{"@type":"Comment","text":" If the graph decreases and increases after passing through the point, then the point will represent the point of minimum. If the graph increases and decreases after passing through the point, then the point will represent the point of maximum."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" The graph has a local maximum in the third quadrant and a local minimum in the 1st quadrant.","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" The graph has a local minimum in the first quadrant and a local maximum in the 1st quadrant.","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" The graph has a local minimum in the third quadrant and a local maximum in the 1st quadrant.","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" The graph has a local minimum in the third quadrant and a local minimum in the 1st quadrant.","position":0,"answerExplanation":{"@type":"Comment","text":"A graph will have a local minimum if the graph decreases before passing through local minimum and increases after passing through local maximum. $$ \\\\ $$ A graph will have a local maximum if the graph increases before passing through local minimum and increases after passing through local maximum. $$ \\\\ $$ From the above graph, it is clear that the local minimum is in the third quadrant and the local maximum is in the first quadrant. ","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Increasing and decreasing intervals Quiz 1","text":" For the function $$f(x)={{x}^{3}}-\\dfrac{{{x}^{2}}}{2}-2x+1$$, find the interval where the function increases?","comment":{"@type":"Comment","text":" A graph decreases between the interval of local maximum and local minimum and increases in the interval between local minimum and local maximum."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\text{(-2,-1)}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\text{(1,2)}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" None of the above","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\left( \\dfrac{\\text{2}}{\\text{3}}\\text{,1} \\right)$$","position":1,"answerExplanation":{"@type":"Comment","text":" $$ f(x)={{x}^{3}}-\\dfrac{{{x}^{2}}}{2}-2x+1 \\\\ \\Rightarrow {f}'\\left( x \\right)=3{{x}^{2}}-x-2 \\\\ \\Rightarrow {f}''(x)=6x-1 \\\\ $$ The values of $$x$$ where $${f}'(x)=0$$ and $${f}''(x) < 0$$, then these $$x$$ values will represent maxima of the function. $$ \\\\ $$ The values of $$x$$ where $${f}'(x)=0$$ and $${f}''(x) > 0$$, then these $$x$$ values will represent minima of the function. $$ \\\\ \\Rightarrow {f}'(x)=0 \\\\ \\Rightarrow 3{{x}^{2}}-x-2=0 \\\\ \\Rightarrow x=\\dfrac{1\\pm \\sqrt{{{1}^{2}}-(4)(3)(-2)}}{6} \\\\ \\Rightarrow x=1,\\dfrac{-2}{3} \\\\ $$ If the value of $$x$$ is equal to $$1$$, then $$ \\\\ f(1)=1-\\dfrac{1}{2}-2+1=\\dfrac{-1}{2} \\\\ {f}''(1)=6(1)-1=5 \\\\ $$ So, at $$x=1$$ we will have local minimum. $$ \\\\ $$ If the value of $$x$$ is equal to $$\\dfrac{\\text{-2}}{\\text{3}}$$, then $$ \\\\ f\\left( \\dfrac{2}{3} \\right)={{\\left( \\dfrac{-2}{3} \\right)}^{3}}-\\dfrac{{{\\left( \\dfrac{2}{3} \\right)}^{2}}}{2}-2\\left( \\dfrac{-2}{3} \\right)+1=\\dfrac{23}{27} \\\\ {f}''\\left( \\dfrac{-2}{3} \\right)=6\\left( \\dfrac{-2}{3} \\right)-1=-5 \\\\ $$ So, if the value of $$x$$ is equal to $$ \\\\ $$ $$\\dfrac{2}{3}$$, we will have local maximum. $$ \\\\ $$ So, it is clear the function will decrease in the interval $$\\left( \\dfrac{2}{3},1 \\right)$$.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Increasing and decreasing intervals Quiz 1","text":" For the function $$f(x)={{x}^{3}}-79{{x}^{2}}+271x-37$$, which of the following statements are correct?","comment":{"@type":"Comment","text":" The values of $x$ where ${f}'(x)=0$ and ${f}''(x) < 0$, then these $x$ values will represent maxima of the function. The values of $x$ where ${f}'(x)=0$ and ${f}''(x) > 0$, then these $x$ values will represent minima of the function."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" The graph will decrease at $$\\left( -\\infty ,1.77 \\right)$$ and increase at $$\\left( 1.77,50.9 \\right)$$. ","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" The graph will increase at $$\\left( 50.9,\\infty \\right)$$ and decrease at $$\\left( 1.77,50.9 \\right)$$.","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" The graph will decrease at $$\\left( -\\infty ,1.77 \\right)\\cup \\left( 50.9,\\infty \\right)$$ and increase at $$\\left( 1.77,50.9 \\right)$$.","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" The graph will increase at $$\\left( -\\infty ,1.77 \\right)\\cup \\left( 50.9,\\infty \\right)$$ and decrease at $$\\left( 1.77,50.9 \\right)$$. ","position":0,"answerExplanation":{"@type":"Comment","text":" $$f(x)={{x}^{3}}-79{{x}^{2}}+270x-37 \\\\ \\Rightarrow {f}'(x)=3{{x}^{2}}-158x+270 \\\\ \\Rightarrow {f}''(x)=6x-158 \\\\ $$ The values of $$x$$ where $${f}'(x)=0$$ and $${f}''(x) < 0$$, then these $$x$$ values will represent maxima of the function. $$ \\\\ $$ The values of $$x$$ where $${f}'(x)=0$$ and $${f}''(x) > 0$$, then these $$x$$ values will represent minima of the function. $$ \\\\ \\Rightarrow {f}'(x)=0 \\\\ \\Rightarrow 3{{x}^{2}}-158x+270=0 \\\\ \\Rightarrow x=\\dfrac{158\\pm \\sqrt{{{158}^{2}}-(4)(3)(270)}}{6} \\\\ \\Rightarrow x=1.77,50.9 \\\\ $$ If the value of $$x$$ is equal to $$1.77$$, then $$ \\\\ \\Rightarrow {f}''(1.77)=6(1.77)-158 < 0 \\\\ f(1.77)={{(1.77)}^{3}}-(79){{(1.77)}^{2}}+270(1.77)-37 > 0 \\\\ $$ So, it is clear that the graph will lie above x-axis $$ \\\\ $$ If the value of $$x$$ is equal to $$50.9$$, then $$ \\\\ \\Rightarrow {f}''(50.9)=6(50.9)-158 > 0 \\\\ f(50.9)={{(50.9)}^{3}}-(79){{(50.9)}^{2}}+270(50.9)-37 < 0 \\\\ $$ So, it is clear that the graph will lie below x-axis. $$ \\\\ $$ So, the function $$f(x)={{x}^{3}}-79{{x}^{2}}+271x-37$$ will have a maximum at the value of $$x$$ equal to $$1.77$$ and minimum at the value of $$x$$ equal to $$50.9$$ $$ \\\\ $$ So, we can say that the graph will be increasing at $$\\left( -\\infty ,1.77 \\right)\\cup \\left( 50.9,\\infty \\right)$$ and will be decreasing at $$\\left( 1.77,50.9 \\right)$$.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}