Let f(x)=sqrt(x+9) and let c be the number that satisfies the Mean Value Theorem for f on the interval [0,16]. What is c ? Choose 1 answer: (A) 16 (B) 43 (C) 55 (D) 7

Question

Let  f(x)=sqrt(x+9) and let  c be the number that satisfies the Mean Value Theorem for  f on the interval  [0,16]. What is  c ? Choose 1 answer: (A) 16 (B) 43 (C) 55 (D) 7

Bytelearn · Accepted Answer

Statement of Theorem: The Mean Value Theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that f'(c) = (f(b) - f(a)) / (b - a). We need to apply this theorem to the function f(x) = sqrt(x+9) on the interval [0, 16].Function Analysis: First, we need to ensure that the function f(x) = sqrt(x+9) is continuous on [0, 16] and differentiable on (0, 16). Since the square root function is continuous and differentiable for all non-negative values of its argument, and x+9 is non-negative for all x in [0, 16], f(x) is indeed continuous on [0, 16] and differentiable on (0, 16).Calculate f(0) and f(16): Next, we calculate f(0) and f(16).  f(0) = sqrt(0+9) = sqrt(9) = 3 f(16) = sqrt(16+9) = sqrt(25) = 5Calculate Average Rate of Change: Now, we calculate the average rate of change of f(x) on the interval [0, 16]: (f(16) - f(0)) / (16 - 0) = (5 - 3) / (16 - 0) = 2 / 16 = 1/8Find Derivative of f(x): To find c, we need to find the value of x for which the instantaneous rate of change, f'(x), is equal to the average rate of change, 1/8. First, we find the derivative of f(x): f'(x) = d/dx [sqrt(x+9)] = 1/(2sqrt(x+9))Set Derivative Equal to Average Rate of Change: We set the derivative equal to the average rate of change and solve for x: 1/(2sqrt(x+9)) = 1/8 Multiplying both sides by 2sqrt(x+9) gives: 1 = 2sqrt(x+9)/8 Multiplying both sides by 8 gives: 8 = 2sqrt(x+9) Dividing both sides by 2 gives: 4 = sqrt(x+9)Solve for x: Squaring both sides to eliminate the square root gives: 16 = x + 9 Subtracting 9 from both sides gives: x = 16 - 9 x = 7Verify c Value: We have found that c = 7 satisfies the Mean Value Theorem for the function f(x) on the interval [0, 16].

Let $f(x)=\sqrt{x+9}$ and let $c$ be the number that satisfies the Mean Value Theorem for $f$ on the interval $[0,16]$ . $\newline$ What is $c$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $16$ $\newline$ (B) $43$ $\newline$ (C) $55$ $\newline$ (D) $7$

Full solution

More problems from Domain and range of square root functions: equations

Related topics

Let f(x)=x+9 f(x)=\sqrt{x+9} f(x)=x+9​ and let c c c be the number that satisfies the Mean Value Theorem for f f f on the interval [0,16] [0,16] [0,16].\newlineWhat is c c c ?\newlineChoose 111 answer:\newline(A) 161616\newline(B) 434343\newline(C) 555555\newline(D) 777

Let $f(x)=\sqrt{x+9}$ and let $c$ be the number that satisfies the Mean Value Theorem for $f$ on the interval $[0,16]$ . $\newline$ What is $c$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $16$ $\newline$ (B) $43$ $\newline$ (C) $55$ $\newline$ (D) $7$