use the intermediate value theorem to show the polynominal function has a zero in the given interval
f(x)=x^5-x^4+3x^3-2x^2-11x+6; [1.5,1.9]
x= -2.33
y=10.19
after i plugged in the 1.5 and 1.9 i just want to know if my x and y are correct
i changed it too
f(1.9)= 10.186
f(1.5)= -2.344
so, the IVT says that if f(x) is continuous, it must take on all values between 10.186 and -2.344.
That includes 0, so there is the proof.
BTW, you values for f are correct.
so i am right? x and y?
x=-2.344
y=10.186
it was continuous, but they wanted it rounded to three decimal places
I wouldn't call your values x and y.
You have two x-values: 1.5 and 1.9
and two values for f(x) or y: -2.344 and 10.186
and, as I said, given those two values for x, your two y values are correct to 3 places.
To use the intermediate value theorem to show that the polynomial function has a zero in the given interval, we need to evaluate the function at the endpoints of the interval and check if the function changes sign.
Let's evaluate the function at the endpoints:
f(1.5) = (1.5)^5 - (1.5)^4 + 3(1.5)^3 - 2(1.5)^2 - 11(1.5) + 6 ≈ -3.69055
f(1.9) = (1.9)^5 - (1.9)^4 + 3(1.9)^3 - 2(1.9)^2 - 11(1.9) + 6 ≈ 4.89145
Now, we have f(1.5) ≈ -3.69055 and f(1.9) ≈ 4.89145. Since the function changes sign from negative to positive between these values, it implies that the function must have at least one zero (root) within the interval [1.5, 1.9].
Now let's check your values:
x = -2.33
y = 10.19
However, the values you provided do not align with the context of the problem. The given interval is [1.5, 1.9], and the values you provided (-2.33, 10.19) are not within that interval. Please check your values again.
Also, keep in mind that using the intermediate value theorem only guarantees the existence of at least one zero in the interval, but it does not provide the exact value or how many zeros there are.