Let f(x)=sqrt(x+sqrt(x^(3))). Below is Lena's attempt to write a formal justification for the fact that the equation f^(')(x)=(2)/(3) has a solution where 0 < x < 9. Is Lena's justification complete? If not, why? Lena's justification: f(0)=0 and f(9)=6, so the average rate of change of f over the interval 0 <= x <= 9 is (2)/(3). So, according to the mean value theorem, f^(')(x)=(2)/(3) must have a solution somewhere in the interval 0 < x < 9. Choose 1 answer: (A) Yes, Lena's justification is complete. (B) No, Lena didn't establish that the average rate of change of f over [0,9] is equal to (2)/(3). (C) No, Lena didn't establish that f is differentiable.

Verify Calculations: Lena states that f(0)=0 and f(9)=6. Let's verify these calculations. f(0) = sqrt(0 + sqrt(0^3)) = sqrt(0 + 0) = sqrt(0) = 0. f(9) = sqrt(9 + sqrt(9^3)) = sqrt(9 + sqrt(729)) = sqrt(9 + 27) = sqrt(36) = 6. These calculations are correct.Calculate Average Rate: Lena calculates the average rate of change of f over the interval [0, 9] as (f(9) - f(0)) / (9 - 0) = (6 - 0) / (9 - 0) = 6 / 9 = 2 / 3. This calculation is also correct.Apply Mean Value Theorem: Lena concludes that since the average rate of change of f over [0, 9] is 2/3, according to the mean value theorem, there must be some c in the interval (0, 9) such that f^(')(c) = 2/3. However, for the mean value theorem to apply, f must be continuous on [0, 9] and differentiable on (0, 9). Lena has not explicitly stated that f is continuous and differentiable on these intervals.Verify Continuity: We need to verify that f is continuous on [0, 9] and differentiable on (0, 9) to apply the mean value theorem. The function f(x) = sqrt(x + sqrt(x^3)) is composed of square root and polynomial functions, which are continuous and differentiable where they are defined. The inner square root, sqrt(x^3), is defined for all x >= 0, and the outer square root, sqrt(x + sqrt(x^3)), is also defined for all x >= 0. Therefore, f is continuous on [0, 9].Check Differentiability: Next, we need to check the differentiability of f on (0, 9). The derivative of a square root function and a polynomial function exists except where the argument of the square root is non-positive. Since the argument of the inner square root, x^3, and the argument of the outer square root, x + sqrt(x^3), are both positive for x > 0, f is differentiable on (0, 9).

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Let
f(x)=sqrt(x+sqrt(x^(3))).
Below is Lena's attempt to write a formal justification for the fact that the equation
f^(')(x)=(2)/(3) has a solution where
0 < x < 9.
Is Lena's justification complete? If not, why?
Lena's justification:

f(0)=0 and
f(9)=6, so the average rate of change of
f over the interval
0 <= x <= 9 is
(2)/(3).
So, according to the mean value theorem,
f^(')(x)=(2)/(3) must have a solution somewhere in the interval
0 < x < 9.
Choose 1 answer:
(A) Yes, Lena's justification is complete.
(B) No, Lena didn't establish that the average rate of change of
f over
[0,9] is equal to
(2)/(3).
(C) No, Lena didn't establish that
f is differentiable.

Let $f(x)=\sqrt{x+\sqrt{x^{3}}}$ . $\newline$ Below is Lena's attempt to write a formal justification for the fact that the equation $f^{\prime}(x)=\frac{2}{3}$ has a solution where \( 0

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Grade 6

One-step inequalities: word problems

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Q. Let

f(x)=\sqrt{x+\sqrt{x^{3}}}

\newline

Below is Lena's attempt to write a formal justification for the fact that the equation

f^{\prime}(x)=\frac{2}{3}

has a solution where

0<x<9

\newline

Is Lena's justification complete? If not, why?

\newline

Lena's justification:

\newline

f(0)=0

and

f(9)=6

, so the average rate of change of

f

over the interval

0 \leq x \leq 9

\frac{2}{3}

\newline

So, according to the mean value theorem,

f^{\prime}(x)=\frac{2}{3}

must have a solution somewhere in the interval

0<x<9

\newline

Choose

1

answer:

\newline

(A) Yes, Lena's justification is complete.

\newline

(B) No, Lena didn't establish that the average rate of change of

f

over

[0,9]

is equal to

\frac{2}{3}

\newline

f

is differentiable.

Verify Calculations: Lena states that $f(0)=0$ and $f(9)=6$ . Let's verify these calculations.f(0) = \sqrt{0 + \sqrt{0^3}} = \sqrt{0 + 0} = \sqrt{0} = 0\.f(9) = \sqrt{9 + \sqrt{9^3}} = \sqrt{9 + \sqrt{729}} = \sqrt{9 + 27} = \sqrt{36} = 6\.These calculations are correct.
Calculate Average Rate: Lena calculates the average rate of change of $f$ over the interval $[0, 9]$ as $(f(9) - f(0)) / (9 - 0) = (6 - 0) / (9 - 0) = 6 / 9 = 2 / 3$ . This calculation is also correct.
Apply Mean Value Theorem: Lena concludes that since the average rate of change of $f$ over $[0, 9]$ is $\frac{2}{3}$ , according to the mean value theorem, there must be some $c$ in the interval $(0, 9)$ such that $f'(c) = \frac{2}{3}$ . However, for the mean value theorem to apply, $f$ must be continuous on $[0, 9]$ and differentiable on $(0, 9)$ . Lena has not explicitly stated that $f$ is continuous and differentiable on these intervals.
Verify Continuity: We need to verify that $f$ is continuous on $[0, 9]$ and differentiable on $(0, 9)$ to apply the mean value theorem. $\newline$ The function $f(x) = \sqrt{x + \sqrt{x^3}}$ is composed of square root and polynomial functions, which are continuous and differentiable where they are defined. $\newline$ The inner square root, $\sqrt{x^3}$ , is defined for all $x \geq 0$ , and the outer square root, $\sqrt{x + \sqrt{x^3}}$ , is also defined for all $x \geq 0$ . $\newline$ Therefore, $f$ is continuous on $[0, 9]$ .
Check Differentiability: Next, we need to check the differentiability of $f$ on $(0, 9)$ . The derivative of a square root function and a polynomial function exists except where the argument of the square root is non-positive. Since the argument of the inner square root, $x^3$ , and the argument of the outer square root, $x + \sqrt{x^3}$ , are both positive for x > 0, $f$ is differentiable on $(0, 9)$ .