Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Let f be a continuous function on the closed interval [-5,0], where f(-5)=0 and f(0)=5. Which of the following is guaranteed by the Intermediate Value Theorem? Choose 1 answer: (A) f(c)=-2 for at least one c between 0 and 5 (B) f(c)=2 for at least one c between 0 and 5 (c) f(c)=-2 for at least one c between -5 and 0 (D) f(c)=2 for at least one c between -5 and 0

Let f f be a continuous function on the closed interval [−5,0] [-5,0] , where f(−5)=0 f(-5)=0 and f(0)=5 f(0)=5 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) f(c)=−2 f(c)=-2 for at least one c c between 00 and 55\newline(B) f(c)=2 f(c)=2 for at least one c c between 00 and 55\newline(C) f(c)=−2 f(c)=-2 for at least one c c between −5-5 and 00\newline(D) f(c)=2 f(c)=2 for at least one c c between −5-5 and 00

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 2right chevron icon
Solve quadratic inequalitiesright chevron icon

Full solution

Q. Let f f be a continuous function on the closed interval [−5,0] [-5,0] , where f(−5)=0 f(-5)=0 and f(0)=5 f(0)=5 .\newlineWhich of the following is guaranteed by the Intermediate Value Theorem?\newlineChoose 11 answer:\newline(A) f(c)=−2 f(c)=-2 for at least one c c between 00 and 55\newline(B) f(c)=2 f(c)=2 for at least one c c between 00 and 55\newline(C) f(c)=−2 f(c)=-2 for at least one c c between −5-5 and 00\newline(D) f(c)=2 f(c)=2 for at least one c c between −5-5 and 00
  1. The Intermediate Value Theorem: The Intermediate Value Theorem states that if ff is a continuous function on a closed interval [a,b][a, b] and NN is any number between f(a)f(a) and f(b)f(b), then there exists at least one cc in the interval (a,b)(a, b) such that f(c)=Nf(c) = N.
  2. Applying the Theorem: Given that ff is continuous on the closed interval [−5,0][-5, 0], f(−5)=0f(-5) = 0, and f(0)=5f(0) = 5, we can apply the Intermediate Value Theorem to find a value cc in the interval (−5,0)(-5, 0) such that f(c)f(c) is any value between 00 and 55.
  3. Evaluating Given Options: We need to determine which of the given options is guaranteed by the Intermediate Value Theorem. Let's evaluate each option:\newline(A) f(c)=−2f(c) = -2 for at least one cc between 00 and 55: This option is not possible because the interval [0,5][0, 5] is not within the domain of ff given by [−5,0][-5, 0].
  4. Option (A): (B) f(c)=2f(c) = 2 for at least one cc between 00 and 55: This option is also not possible for the same reason as option (A); the interval [0,5][0, 5] is not within the domain of ff given by [−5,0][-5, 0].
  5. Option (B): C)f(c)=−2C) f(c) = -2 for at least one cc between \$\(-5\)\) and \$\(0\)\): This option is not possible because the values of \(f\) on the interval \[{-5, 0}\] are between \$\(0\)\) and \$\(5\)\), and \$\(-2\)\) is not between these values.
  6. Option (C): (D) \(f(c) = 2\) for at least one \(c\) between \(-5\) and \(0\): This option is possible because \(2\) is a value between \(f(-5) = 0\) and \(f(0) = 5\). By the Intermediate Value Theorem, there must be at least one \(c\) in the interval \((-5, 0)\) such that \(f(c) = 2\).

More problems from Solve quadratic inequalities

Question
−10x−3≥8x+4 -10 x-3 \geq 8 x+4 \newlineWhich of the following best describes the solutions to the inequality shown?\newlineChoose 11 answer:\newlineA) x≤−718 x \leq-\frac{7}{18} \newline(B) x≤−72 x \leq-\frac{7}{2} \newline(C) x≤−118 x \leq-\frac{1}{18} \newline(D) x≥−718 x \geq-\frac{7}{18}
Get tutor helpright-arrow

Posted 10 months ago

Question
−12x<−72 -12 x<-72 \newlineWhich of the following best describes the solutions to the inequality shown?\newlineChoose 11 answer:\newline(A) x<6 x<6 \newline(B) x>6 x>6 \newline(C) x<16 x<\frac{1}{6} \newline(D) x>16 x>\frac{1}{6}
Get tutor helpright-arrow

Posted 10 months ago

Question
8x+6<4x+10 8 x+6<4 x+10 \newlineWhich of the following best describes the solutions to the inequality shown?\newlineChoose 11 answer:\newline(A) x<13 x<\frac{1}{3} \newline(B) x<4 x<4 \newline(C) x<1 x<1 \newline(D) x>1 x>1
Get tutor helpright-arrow

Posted 10 months ago

Question
6x−2<2x+3 6 x-2<2 x+3 \newlineWhich of the following best describes the solutions to the inequality shown?\newlineChoose 11 answer:\newline(A) x<54 x<\frac{5}{4} \newline(B) x>54 x>\frac{5}{4} \newline(C) x<14 x<\frac{1}{4} \newline(D) x<58 x<\frac{5}{8}
Get tutor helpright-arrow

Posted 10 months ago

Question
16<−4x 16<-4 x \newlineWhich of the following best describes the solutions to the inequality shown?\newlineChoose 11 answer:\newline(A) x<−4 x<-4 \newline(B) x>−4 x>-4 \newline(C) x<−14 x<-\frac{1}{4} \newline(D) x>−14 x>-\frac{1}{4}
Get tutor helpright-arrow

Posted 10 months ago

Question
Find lim⁡x→42−4x−12x−4 \lim _{x \rightarrow 4} \frac{2-\sqrt{4 x-12}}{x-4} .\newlineChoose 11 answer:\newline(A) 22\newline(B) 11\newline(C) −1-1\newline(D) The limit doesn't exist
Get tutor helpright-arrow

Posted 9 months ago

Question
Find lim⁡x→2x4−4x3+4x2x−2 \lim _{x \rightarrow 2} \frac{x^{4}-4 x^{3}+4 x^{2}}{x-2} .\newlineChoose 11 answer:\newline(A) −4-4\newline(B) 00\newline(C) 44\newline(D) The limit doesn't exist
Get tutor helpright-arrow

Posted 9 months ago

Question
Let g(x)=tan⁡(x) g(x)=\tan (x) .\newlineCan we use the intermediate value theorem to say the equation g(x)=0 g(x)=0 has a solution where π4≤x≤3π4 \frac{\pi}{4} \leq x \leq \frac{3 \pi}{4} ?\newlineChoose 11 answer:\newline(A) No, since the function is not continuous on that interval.\newline(B) No, since 00 is not between g(π4) g\left(\frac{\pi}{4}\right) and g(3π4) g\left(\frac{3 \pi}{4}\right) .\newline(C) Yes, both conditions for using the intermediate value theorem have been met.
Get tutor helpright-arrow

Posted 9 months ago

Question
Let g(x)=cos⁡(x) g(x)=\cos (x) .\newlineCan we use the intermediate value theorem to say the equation g(x)=0.8 g(x)=0.8 has a solution where 0≤x≤π2 0 \leq x \leq \frac{\pi}{2} ?\newlineChoose 11 answer:\newline(A) No, since the function is not continuous on that interval.\newline(B) No, since 00.88 is not between g(0) g(0) and g(π2) g\left(\frac{\pi}{2}\right) .\newline(C) Yes, both conditions for using the intermediate value theorem have been met.
Get tutor helpright-arrow

Posted 9 months ago

Question
You pick a card at random, put it back, and then pick another card at random.\newline44\newline55\newline66\newline77\newlineWhat is the probability of picking a number greater than 55 and then picking a 44 ?\newlineWrite your answer as a percentage.
Get tutor helpright-arrow

Posted 1 month ago

View all (25)

Related topics

Algebra - Order of OperationsAlgebra - Distributive Property`X` and `Y` AxesGeometry - Scalene TriangleCommon MultipleGeometry - Quadrant