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Let 
g(x)=sqrt(5x-1) and let 
c be the number that satisfies the Mean Value Theorem for 
g on the interval 
[1,10].
What is 
c ?
Choose 1 answer:
(A) 2.25
(B) 4.25
(C) 6.5
(D) 8

Let g(x)=5x1 g(x)=\sqrt{5 x-1} and let c c be the number that satisfies the Mean Value Theorem for g g on the interval [1,10] [1,10] .\newlineWhat is c c ?\newlineChoose 11 answer:\newline(A) 22.2525\newline(B) 44.2525\newline(C) 66.55\newline(D) 88

Full solution

Q. Let g(x)=5x1 g(x)=\sqrt{5 x-1} and let c c be the number that satisfies the Mean Value Theorem for g g on the interval [1,10] [1,10] .\newlineWhat is c c ?\newlineChoose 11 answer:\newline(A) 22.2525\newline(B) 44.2525\newline(C) 66.55\newline(D) 88
  1. Define Mean Value Theorem: The Mean Value Theorem states that if a function is continuous on the closed interval [a,b][a, b] and differentiable on the open interval (a,b)(a, b), then there exists at least one number cc in the interval (a,b)(a, b) such that g(c)=g(b)g(a)bag'(c) = \frac{g(b) - g(a)}{b - a}. Let's calculate g(x)g'(x) first.
  2. Calculate g(x)g'(x): To find g(x)g'(x), we need to differentiate g(x)=5x1g(x) = \sqrt{5x - 1}. Using the chain rule, g(x)=(12)(5x1)125g'(x) = (\frac{1}{2}) \cdot (5x - 1)^{-\frac{1}{2}} \cdot 5.
  3. Find Change in gg: Now we simplify g(x)g'(x) to get g(x)=525x1g'(x) = \frac{5}{2 \cdot \sqrt{5x - 1}}.
  4. Find Change in x: Next, we calculate g(10)g(10) and g(1)g(1) to find the change in gg over the interval [1,10][1, 10]. g(10)=5×101=49=7g(10) = \sqrt{5\times10 - 1} = \sqrt{49} = 7 and g(1)=5×11=4=2g(1) = \sqrt{5\times1 - 1} = \sqrt{4} = 2.
  5. Apply Mean Value Theorem: Now we find the change in gg over the interval [1,10][1, 10], which is g(10)g(1)=72=5g(10) - g(1) = 7 - 2 = 5.
  6. Set up Equation: We also find the change in xx over the interval [1,10][1, 10], which is 101=910 - 1 = 9.
  7. Cross-Multiply: Now we can set up the equation from the Mean Value Theorem: g(c)=g(10)g(1)101g'(c) = \frac{g(10) - g(1)}{10 - 1}. Substituting the values we have, we get 525c1=59\frac{5}{2 \cdot \sqrt{5c - 1}} = \frac{5}{9}.
  8. Solve for c: We can now solve for c by cross-multiplying: 5×9=2×5c1×55 \times 9 = 2 \times \sqrt{5c - 1} \times 5.
  9. Eliminate Square Root: Simplifying the equation, we get 45=10×5c145 = 10 \times \sqrt{5c - 1}.
  10. Calculate cc: Dividing both sides by 1010, we get 4.5=5c14.5 = \sqrt{5c - 1}.
  11. Calculate c: Dividing both sides by 1010, we get 4.5=5c14.5 = \sqrt{5c - 1}. Squaring both sides to eliminate the square root, we get 4.52=(5c1)4.5^2 = (5c - 1).
  12. Calculate c: Dividing both sides by 1010, we get 4.5=5c14.5 = \sqrt{5c - 1}. Squaring both sides to eliminate the square root, we get 4.52=(5c1)4.5^2 = (5c - 1). Calculating 4.524.5^2 gives us 20.25=5c120.25 = 5c - 1.
  13. Calculate c: Dividing both sides by 1010, we get 4.5=5c14.5 = \sqrt{5c - 1}. Squaring both sides to eliminate the square root, we get 4.52=(5c1)4.5^2 = (5c - 1). Calculating 4.524.5^2 gives us 20.25=5c120.25 = 5c - 1. Adding 11 to both sides, we get 20.25+1=5c20.25 + 1 = 5c.
  14. Calculate c: Dividing both sides by 1010, we get 4.5=5c14.5 = \sqrt{5c - 1}. Squaring both sides to eliminate the square root, we get 4.52=(5c1)4.5^2 = (5c - 1). Calculating 4.524.5^2 gives us 20.25=5c120.25 = 5c - 1. Adding 11 to both sides, we get 20.25+1=5c20.25 + 1 = 5c. Simplifying, we get 21.25=5c21.25 = 5c.
  15. Calculate c: Dividing both sides by 1010, we get 4.5=5c14.5 = \sqrt{5c - 1}. Squaring both sides to eliminate the square root, we get 4.52=(5c1)4.5^2 = (5c - 1). Calculating 4.524.5^2 gives us 20.25=5c120.25 = 5c - 1. Adding 11 to both sides, we get 20.25+1=5c20.25 + 1 = 5c. Simplifying, we get 21.25=5c21.25 = 5c. Finally, dividing both sides by 55, we find c=21.255=4.25c = \frac{21.25}{5} = 4.25.

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