Use the intermediate value theorem to show that f(x) has a zero in the given interval. Please show all of your work.
f(x) = 3x^3 + 8x^2 - 5x - 11; [-2.8, -2.7]
I'm not understanding all this questions...
If the function f(x) is continuous on
[a,b] and f(a)*f(b)<0 then on (a,b) there
are at least one zero. Polynomial is
continuous on R.
Compute f(-2.8) and f(-2.7)
No problem! I will explain the intermediate value theorem and show you how to apply it to determine if the function has a zero in the given interval.
The intermediate value theorem states that if a function is continuous on a closed interval [a, b] and takes on two different values f(a) and f(b), then it must take on every value between f(a) and f(b) at least once on the interval.
In other words, if a function is continuous and goes from a negative value to a positive value (or vice versa) on an interval, it must cross the x-axis and have a zero in that interval.
Now let's apply the intermediate value theorem to the function f(x) = 3x^3 + 8x^2 - 5x - 11 on the interval [-2.8, -2.7].
Step 1: Calculate the values of the function at both endpoints of the interval.
When x = -2.8:
f(-2.8) = 3(-2.8)^3 + 8(-2.8)^2 - 5(-2.8) - 11 = -49.872
When x = -2.7:
f(-2.7) = 3(-2.7)^3 + 8(-2.7)^2 - 5(-2.7) - 11 = -24.609
Step 2: Check if the function changes sign between the endpoint values.
Since f(-2.8) = -49.872 and f(-2.7) = -24.609, we can see that the function changes sign between these values.
Step 3: Conclusion
By the intermediate value theorem, since the function changes sign between the values of f(-2.8) and f(-2.7), there must be at least one zero in the interval [-2.8, -2.7].
So, f(x) has a zero in the given interval.