Let f(x)=2^(5-2x). Below is Ibrahim's attempt to write a formal justification for the fact that the equation f(x)=10 has a solution where -1 <= x <= 4. Is Ibrahim's justification complete? If not, why? Ibrahim's justification: f is defined for all real numbers, and exponential functions are continuous at all points in their domains. Also, f(-1)=128 and f(4)=(1)/(8), so 10 is between f(-1) and f(4). So, according to the intermediate value theorem, f(x)=10 must have a solution at some point between x=-1 and x=4. Choose 1 answer: (A) Yes, Ibrahim's justification is complete. (B) No, Ibrahim didn't establish that 10 is between f(-1) and f(4). (c) No, Ibrahim didn't establish that f is continuous.

Function Definition: Ibrahim states that the function f is defined for all real numbers, which is true for exponential functions like f(x) = 2^(5-2x). This means that the function does not have any points of discontinuity.Continuity of Exponential Functions: Ibrahim also states that exponential functions are continuous at all points in their domains. This is a correct statement, as exponential functions do not have breaks, holes, or jumps in their graphs.Calculation of f(-1): Ibrahim calculates f(-1) = 2^(5 - 2(-1)) = 2^(5 + 2) = 2^7 = 128. This calculation is correct.Calculation of f(4): Ibrahim calculates f(4) = 2^(5 - 2(4)) = 2^(5 - 8) = 2^(-3) = 1/2^3 = 1/8. This calculation is also correct.Application of Intermediate Value Theorem: Ibrahim then states that since 10 is between f(-1) = 128 and f(4) = 1/8, by the intermediate value theorem, there must be a solution to f(x) = 10 somewhere between x = -1 and x = 4. However, Ibrahim has not explicitly shown that 10 is between f(-1) and f(4). To complete the justification, Ibrahim needs to show that f(-1) > 10 and f(4) 10 and 1/8 < 10. Therefore, 10 is indeed between f(-1) and f(4), and Ibrahim's justification is now complete with this additional information.

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Let
f(x)=2^(5-2x).
Below is Ibrahim's attempt to write a formal justification for the fact that the equation
f(x)=10 has a solution where
-1 <= x <= 4.
Is Ibrahim's justification complete? If not, why?
Ibrahim's justification:

f is defined for all real numbers, and exponential functions are continuous at all points in their domains. Also,
f(-1)=128 and
f(4)=(1)/(8), so 10 is between
f(-1) and
f(4).
So, according to the intermediate value theorem,
f(x)=10 must have a solution at some point between
x=-1 and
x=4.
Choose 1 answer:
(A) Yes, Ibrahim's justification is complete.
(B) No, Ibrahim didn't establish that 10 is between
f(-1) and
f(4).
(c) No, Ibrahim didn't establish that
f is continuous.

Let $f(x)=2^{5-2 x}$ . $\newline$ Below is Ibrahim's attempt to write a formal justification for the fact that the equation $f(x)=10$ has a solution where $-1 \leq x \leq 4$ . $\newline$ Is Ibrahim's justification complete? If not, why? $\newline$ Ibrahim's justification: $\newline$ $f$ is defined for all real numbers, and exponential functions are continuous at all points in their domains. Also, $f(-1)=128$ and $f(4)=\frac{1}{8}$ , so $10$ is between $f(-1)$ and $f(4)$ . $\newline$ So, according to the intermediate value theorem, $f(x)=10$ must have a solution at some point between $x=-1$ and $x=4$ . $\newline$ Choose $1$ answer: $\newline$ (A) Yes, Ibrahim's justification is complete. $\newline$ (B) No, Ibrahim didn't establish that $10$ is between $f(-1)$ and $f(4)$ . $\newline$ (c) No, Ibrahim didn't establish that $f$ is continuous.

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Math Problems

Grade 7

One-step inequalities: word problems

Full solution

Q. Let

f(x)=2^{5-2 x}

\newline

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

f(x)=10

has a solution where

-1 \leq x \leq 4

\newline

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

\newline

Ibrahim's justification:

\newline

f

is defined for all real numbers, and exponential functions are continuous at all points in their domains. Also,

f(-1)=128

and

f(4)=\frac{1}{8}

, so

10

is between

f(-1)

and

f(4)

\newline

So, according to the intermediate value theorem,

f(x)=10

must have a solution at some point between

x=-1

and

x=4

\newline

Choose

1

answer:

\newline

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

\newline

(B) No, Ibrahim didn't establish that

10

is between

f(-1)

and

f(4)

\newline

f

is continuous.

Function Definition: Ibrahim states that the function $f$ is defined for all real numbers, which is true for exponential functions like $f(x) = 2^{(5-2x)}$ . This means that the function does not have any points of discontinuity.
Continuity of Exponential Functions: Ibrahim also states that exponential functions are continuous at all points in their domains. This is a correct statement, as exponential functions do not have breaks, holes, or jumps in their graphs.
Calculation of $f(-1)$ : Ibrahim calculates $f(-1) = 2^{5 - 2(-1)} = 2^{5 + 2} = 2^7 = 128$ . This calculation is correct.
Calculation of $f(4)$ : Ibrahim calculates $f(4) = 2^{5 - 2(4)} = 2^{5 - 8} = 2^{-3} = \frac{1}{2^3} = \frac{1}{8}$ . This calculation is also correct.
Application of Intermediate Value Theorem: Ibrahim then states that since $10$ is between $f(-1) = 128$ and $f(4) = \frac{1}{8}$ , by the intermediate value theorem, there must be a solution to $f(x) = 10$ somewhere between $x = -1$ and $x = 4$ . However, Ibrahim has not explicitly shown that $10$ is between $f(-1)$ and $f(4)$ . To complete the justification, Ibrahim needs to show that f(-1) > 10 and $f(-1) = 128$ $0$ .
Justification of Intermediate Value Theorem: Since $f(-1) = 128$ and $f(4) = \frac{1}{8}$ , we can see that 128 > 10 and \frac{1}{8} < 10. Therefore, $10$ is indeed between $f(-1)$ and $f(4)$ , and Ibrahim's justification is now complete with this additional information.