Answer:
the answer would be c.
Step-by-step explanation:
Answer:
See verification below
Step-by-step explanation:
Remember that the Intermediate Value Theorem (IVT) states that if f is a continuous function on [a,b] and f(a)<M<f(b), there exists some c∈[a,b] such that f(c)=M.
Now, to apply the theorem, we have that f(0)=0²+0-1=-1, f(5)=5²+5-1=29, then f(0)=-1<11<29=f(5). Additionally, f is continous since it is a polynomial. Then the IVT applies, and such c exists.
To find, c, solve the quadratic equation f(c)=11. This equation is c²+c-1=11. Rearranging, c²+c-12=0. Factor the expression to get (c+4)(c-3)=0. Then c=-4 or c=3. -4 is not in the interval, then we take c=3. Indeed, f(3)=3²+3-1=9+3-1=11.
Answer:
Answer c: 1000
Step-by-step explanation:
Set f(x) = 2 log x - 6 = 0 as a first step in finding the zero or zeros.
2 log x - 6 = 0 becomes log x - 3 = 0 after dividing all three terms by 2.
Then log x = 3, and because "log x" has base 10, the value of x is 10^3, or 1000.
