Let f(x)=root(3)(cos(pi*x)). Below is Issac's attempt to write a formal justification for the fact that the equation f(x)=0.1 has a solution where 1.5 <= x <= 2. Is Issac's justification complete? If not, why? Issac's justification: {:[f(1.5)=0" and "],[f(2)=1", so "],[f(1.5) < 0.1 < f(2).]:} So, according to the intermediate value theorem, f(x)=0.1 must have a solution when x is between x=1.5 and x=2. Choose 1 answer: (A) Yes, Issac's justification is complete. (B) No, Issac didn't establish that 0.1 is between f(1.5) and f(2). (C) No, Issac didn't establish that f is continuous.

Check Continuity: To apply the Intermediate Value Theorem (IVT), we need to ensure that the function f(x) is continuous on the interval [1.5, 2] and that the value 0.1 lies between f(1.5) and f(2).Verify Function Values: First, we need to check if f(x) is continuous on the interval [1.5, 2]. The function f(x) = root(3)(cos(pi*x)) is continuous because the cube root function and the cosine function are both continuous everywhere.Calculate f(1.5): Next, we need to verify Issac's claims about the values of f(1.5) and f(2). We calculate f(1.5) and f(2) to ensure that 0.1 is indeed between these two values.Calculate f(2): Calculate f(1.5) = root(3)(cos(pi*1.5)). Since cos(pi*1.5) = cos(3pi/2) = 0, we have f(1.5) = root(3)(0) = 0.Confirm Value Range: Calculate f(2) = root(3)(cos(pi*2)). Since cos(pi*2) = cos(2pi) = 1, we have f(2) = root(3)(1) = 1.Apply IVT: Now we can see that 0.1 is indeed between f(1.5) = 0 and f(2) = 1. Therefore, 0 < 0.1 < 1.Since f(x) is continuous on [1.5, 2] and 0.1 is between f(1.5) and f(2), by the Intermediate Value Theorem, there must be some value c in [1.5, 2] such that f(c) = 0.1.

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Let
f(x)=root(3)(cos(pi*x)).
Below is Issac's attempt to write a formal justification for the fact that the equation
f(x)=0.1 has a solution where
1.5 <= x <= 2.
Is Issac's justification complete? If not, why?
Issac's justification:

{:[f(1.5)=0" and "],[f(2)=1", so "],[f(1.5) < 0.1 < f(2).]:}
So, according to the intermediate value theorem,
f(x)=0.1 must have a solution when
x is between
x=1.5 and
x=2.
Choose 1 answer:
(A) Yes, Issac's justification is complete.
(B) No, Issac didn't establish that 0.1 is between
f(1.5) and
f(2).
(C) No, Issac didn't establish that
f is continuous.

Let $f(x)=\sqrt[3]{\cos (\pi \cdot x)}$ . $\newline$ Below is Issac's attempt to write a formal justification for the fact that the equation $f(x)=0.1$ has a solution where $1.5 \leq x \leq 2$ . $\newline$ Is Issac's justification complete? If not, why? $\newline$ Issac's justification: $\newline$ \[ \begin{array}{l} f(1.5)=0 \text { and } \\ f(2)=1 \text {, so } \\ f(1.5)<0.1

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Grade 7

One-step inequalities: word problems

Full solution

Q. Let

f(x)=\sqrt[3]{\cos (\pi \cdot x)}

\newline

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

f(x)=0.1

has a solution where

1.5 \leq x \leq 2

\newline

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

\newline

Issac's justification:

\newline

\begin{array}{l} f(1.5)=0 \text { and } \\ f(2)=1 \text {, so } \\ f(1.5)<0.1<f(2) . \end{array}

\newline

So, according to the intermediate value theorem,

f(x)=0.1

must have a solution when

x

is between

x=1.5

and

x=2

\newline

Choose

1

answer:

\newline

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

\newline

(B) No, Issac didn't establish that

0

1

is between

f(1.5)

and

f(2)

\newline

f

is continuous.

Check Continuity: To apply the Intermediate Value Theorem (IVT), we need to ensure that the function $f(x)$ is continuous on the interval $[1.5, 2]$ and that the value $0.1$ lies between $f(1.5)$ and $f(2)$ .
Verify Function Values: First, we need to check if $f(x)$ is continuous on the interval $[1.5, 2]$ . The function $f(x) = \sqrt[3]{\cos(\pi x)}$ is continuous because the cube root function and the cosine function are both continuous everywhere.
Calculate $f(1.5)$ : Next, we need to verify Issac's claims about the values of $f(1.5)$ and $f(2)$ . We calculate $f(1.5)$ and $f(2)$ to ensure that $0.1$ is indeed between these two values.
Calculate $f(2)$ : Calculate $f(1.5) = \sqrt[3]{\cos(\pi\cdot1.5)}$ . Since $\cos(\pi\cdot1.5) = \cos(\frac{3\pi}{2}) = 0$ , we have $f(1.5) = \sqrt[3]{0} = 0$ .
Confirm Value Range: Calculate $f(2) = \sqrt[3]{\cos(\pi\cdot 2)}$ . Since $\cos(\pi\cdot 2) = \cos(2\pi) = 1$ , we have $f(2) = \sqrt[3]{1} = 1$ .
Apply IVT: Now we can see that $0.1$ is indeed between $f(1.5) = 0$ and $f(2) = 1$ . Therefore, 0 < 0.1 < 1.
Apply IVT: Now we can see that $0.1$ is indeed between $f(1.5) = 0$ and $f(2) = 1$ . Therefore, 0 < 0.1 < 1.Since $f(x)$ is continuous on $[1.5, 2]$ and $0.1$ is between $f(1.5)$ and $f(2)$ , by the Intermediate Value Theorem, there must be some value $c$ in $[1.5, 2]$ such that $f(1.5) = 0$ $1$ .