Maths help ?
I tried to answer this q by subbing in (-2) , and (-1) but still go the wrong answer could someone please explain
- PopeLv 72 months ago
You must be referring to part (a). What did you get when you evaluated the function at -2 and -1? Is one negative and the other positive? The function is a polynomial, hence continuous an all teal numbers. It cannot move between negative and positive without meeting zero somewhere on the way.
- ted sLv 72 months ago
a) f(-2) > 0 while f(-1) < 0 and f is continuous ====> it crosses the x axis
c)f ' ( x) = 4 ( x³ - 1)====> x = 1 is a critical value
b)...divide f(x) by ' x - 2 ' to find a , b , c.....not hard
d) 4th degree means starts high and ends high...using parts b & c we can deduce that a 2nd root exists...function decreases for x < 1 and increases for x > 1
- Anonymous2 months ago
f(x) = x^4 - 4x - 8 = 16 for x = -2. the function = -3 for x = -1. Since you go from 16 to -3, the function must equal 0 at some point between those two numbers and that is a root. Specifically, about x = -1.296