Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 14

MATH100 A December 2023

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[hide] Question 14
Let ${\textstyle f(x)=x^{4}-4x^{3}-18x^{2}+13}$. Find all intervals where ${\textstyle f(x)}$ is concave up.

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[show] Hint
Recall that the second derivative tells you information about the concavity of a function.

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[show] Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Recall how the second derivative gives us information about the concavity of a function: a function $f$ is concave up wherever the second derivative $f''$ is positive and concave down wherever the second derivative is negative. Thus the concavity changes at the points where the second derivative vanishes. First, let’s compute some derivatives: $${\begin{aligned}f'(x)&=4x^{3}-12x^{2}-36x,\\f''(x)&=12x^{2}-24x-36.\end{aligned}}$$ Now, set the second derivative to zero and solve for the points where the concavity changes: $${\begin{aligned}f''(x)=12x^{2}-24x-36&=0\\12(x^{2}-2x-3)&=0\\(x-3)(x+1)&=0.\end{aligned}}$$ So, we now have three intervals: ${\textstyle (-\infty ,-1)}$, ${\textstyle (-1,3)}$, and ${\textstyle (3,\infty )}$, and we need to determine in which ones the function is concave up. We can do this by evaluating the second derivative at a point in the interval to see if it’s positive or negative. ${\textstyle f''(-2)=60>0}$, ${\textstyle f''(0)=-36<0}$, and ${\textstyle f''(4)=60>0}$, so we can conclude that ${\textstyle f(x)}$ is concave up over the intervals ${\textstyle (-\infty ,-1)}$ and ${\textstyle (3,\infty )}$.