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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)=252x f(x)=2^{5-2 x} .\newlineBelow is Ibrahim's attempt to write a formal justification for the fact that the equation f(x)=10 f(x)=10 has a solution where 1x4 -1 \leq x \leq 4 .\newlineIs Ibrahim's justification complete? If not, why?\newlineIbrahim's justification:\newlinef f is defined for all real numbers, and exponential functions are continuous at all points in their domains. Also, f(1)=128 f(-1)=128 and f(4)=18 f(4)=\frac{1}{8} , so 1010 is between f(1) f(-1) and f(4) f(4) .\newlineSo, according to the intermediate value theorem, f(x)=10 f(x)=10 must have a solution at some point between x=1 x=-1 and x=4x=4 .\newlineChoose 11 answer:\newline(A) Yes, Ibrahim's justification is complete.\newline(B) No, Ibrahim didn't establish that 1010 is between f(1) f(-1) and f(4) f(4) .\newline(c) No, Ibrahim didn't establish that f f is continuous.

Full solution

Q. Let f(x)=252x f(x)=2^{5-2 x} .\newlineBelow is Ibrahim's attempt to write a formal justification for the fact that the equation f(x)=10 f(x)=10 has a solution where 1x4 -1 \leq x \leq 4 .\newlineIs Ibrahim's justification complete? If not, why?\newlineIbrahim's justification:\newlinef f is defined for all real numbers, and exponential functions are continuous at all points in their domains. Also, f(1)=128 f(-1)=128 and f(4)=18 f(4)=\frac{1}{8} , so 1010 is between f(1) f(-1) and f(4) f(4) .\newlineSo, according to the intermediate value theorem, f(x)=10 f(x)=10 must have a solution at some point between x=1 x=-1 and x=4x=4 .\newlineChoose 11 answer:\newline(A) Yes, Ibrahim's justification is complete.\newline(B) No, Ibrahim didn't establish that 1010 is between f(1) f(-1) and f(4) f(4) .\newline(c) No, Ibrahim didn't establish that f f is continuous.
  1. Function Definition: Ibrahim states that the function ff is defined for all real numbers, which is true for exponential functions like f(x)=2(52x)f(x) = 2^{(5-2x)}. This means that the function does not have any points of discontinuity.
  2. 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.
  3. Calculation of f(1)f(-1): Ibrahim calculates f(1)=252(1)=25+2=27=128f(-1) = 2^{5 - 2(-1)} = 2^{5 + 2} = 2^7 = 128. This calculation is correct.
  4. Calculation of f(4)f(4): Ibrahim calculates f(4)=252(4)=258=23=123=18f(4) = 2^{5 - 2(4)} = 2^{5 - 8} = 2^{-3} = \frac{1}{2^3} = \frac{1}{8}. This calculation is also correct.
  5. Application of Intermediate Value Theorem: Ibrahim then states that since 1010 is between f(1)=128f(-1) = 128 and f(4)=18f(4) = \frac{1}{8}, by the intermediate value theorem, there must be a solution to f(x)=10f(x) = 10 somewhere between x=1x = -1 and x=4x = 4. However, Ibrahim has not explicitly shown that 1010 is between f(1)f(-1) and f(4)f(4). To complete the justification, Ibrahim needs to show that f(-1) > 10 and f(1)=128f(-1) = 12800.
  6. Justification of Intermediate Value Theorem: Since f(1)=128f(-1) = 128 and f(4)=18f(4) = \frac{1}{8}, we can see that 128 > 10 and \frac{1}{8} < 10. Therefore, 1010 is indeed between f(1)f(-1) and f(4)f(4), and Ibrahim's justification is now complete with this additional information.

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