If f(x) = -1/x, then f'(x) = 1/x^2. Theorem seems to suggest that the integral from -1 to 1 of 1/x^2 dx would equal f(1) - f(-1) = -1 -1 = -2. But 1/x^2 is a positive function and so its integral over [-1,1] should be positive. What is wrong here?
Answers
Answer:
The flaw is at x=0 where the function is not defined
Step-by-step explanation:
For any function we have
The integral can be evaluated only if the function f(x) is defined in the interval [a,b] in which this integral is evaluated
In our our case function f(x) is not defined at x=0 in the interval [-1,+1] thus this results in the flaw that we have obtained in the reasoning
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Step-by-step explanation: i got it right on my assignment
Answers
Answer: x = 0
Explanation
The idea is to replace h(x) with 3. Then we isolate x. Follow PEMDAS in reverse to get x by itself.
Here are the steps:
h(x) = -5x+3
3 = -5x+3
-5x+3 = 3
-5x = 3-3
-5x = 0
x = 0/(-5)
x = 0
As a check,
h(x) = -5x+3
h(0) = -5*0 + 3
h(0) = 0 + 3
h(0) = 3
Therefore, the input x = 0 in the domain leads to the output h(x) = 3 in the range. We have confirmed the answer x = 0
Another way to confirm the answer is to use a graphing tool like GeoGebra or Desmos.