Why would there be a change of sign since there isnt a root?
There isn’t a root as such, if there are solutions to f(x) = b, over a small interval, subbing in two values of x either side of the root would produce a change of sign of f(x) - b = 0, as long as the function is continuous.
There isn’t a root as such, if there are solutions to f(x) = b, over a small interval, subbing in two values of x either side of the root would produce a change of sign of f(x) - b = 0, as long as the function is continuous.
Think back to what you have learnt in iteration
George Kambosos Jr.
Oh yh i wasnt taught this topic
Still slightly confused about why it would produce a change of sign cause there isn’t a root, but ig ill just think about it
My bad, for some reason I read that symbol as ‘root’. Maybe the sign change is down to a change in gradient. Best thing you can do at this stage is just learn and memorize how things are done. That’s what I do.
It would probably be better for you to try and sketch whats happening. * The function is increasing. * You know (can evaluate) 1.5^1.5 < 2 < 1.6^1.6 or f(1.5) < f(x) < f(1.6) So what can you conclude about the point where f(x)=2?
It would probably be better for you to try and sketch whats happening. * The function is increasing. * You know (can evaluate) 1.5^1.5 < 2 < 1.6^1.6 or f(1.5) < f(x) < f(1.6) So what can you conclude about the point where f(x)=2?
Of course. The first part asked you to find the location of hte minimum (1/e) and to the right of the minimum, the function is increasing. You use that knowledge in this part to show the function must be equal to 2 somewhere between 1.5 and 1.6.
Of course. The first part asked you to find the location of hte minimum (1/e) and to the right of the minimum, the function is increasing. You use that knowledge in this part to show the function must be equal to 2 somewhere between 1.5 and 1.6.
That makes sense but why is the change of sign relevant in this case? Thank u
That makes sense but why is the change of sign relevant in this case? Thank u
Its similar to a root finding problem. Consider the function g(x) = f(x)-2 as mentioned above. Then g(1.5) < 0 g(1.6) > 0 so g() has a root somewhere between 1.5 and 1.6 which is the location where f()=2. Its the same argument.
Its similar to a root finding problem. Consider the function g(x) = f(x)-2 as mentioned above. Then g(1.5) < 0 g(1.6) > 0 so g() has a root somewhere between 1.5 and 1.6 which is the location where f()=2. Its the same argument.
Its a function. Have you sketched what you think is happening - can you upload? In general, your responses are very terse and it would help for your replies to contain more content about what you do undersetand ...