[amp_mcq option1=”18″ option2=”10″ option3=”-2.25″ option4=”indeterminate” correct=”option1″]<\/p>\n
The maximum value of $f(x) = x^2 – x – 2$ in the closed interval $[-4, 4]$ is $\\boxed{10}$.<\/p>\n
To find the maximum value, we can first find the critical points of $f$. The derivative of $f$ is $f'(x) = 2x – 1$. Setting $f'(x) = 0$, we find that the critical point is $x = \\frac{1}{2}$.<\/p>\n
Since $f$ is a polynomial, it is continuous for all real numbers. Therefore, according to the Extreme Value Theorem, $f$ must have an absolute maximum (or minimum) over any closed interval.<\/p>\n
To find the absolute maximum value of $f$, we can evaluate $f$ at the critical point and at the endpoints of the interval. The critical point is $x = \\frac{1}{2}$, and the endpoints of the interval are $x = -4$ and $x = 4$. Therefore, the possible maximum values of $f$ are $f\\left(\\frac{1}{2}\\right), f(-4)$, and $f(4)$.<\/p>\n
We can evaluate $f$ at each of these points to find that $f\\left(\\frac{1}{2}\\right) = \\frac{9}{4} < 10$, $f(-4) = 16 – 4 – 2 = 10$, and $f(4) = 16 – 4 – 2 = 10$. Therefore, the maximum value of $f$ is $\\boxed{10}$.<\/p>\n
The other options are incorrect because they are not the maximum value of $f$. Option A is incorrect because $18 > 10$. Option B is incorrect because $10$ is the maximum value of $f$. Option C is incorrect because $-2.25 < 0$. Option D is incorrect because the maximum value of $f$ is not indeterminate.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”18″ option2=”10″ option3=”-2.25″ option4=”indeterminate” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20194","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n