Show that the equation x^3+6x-5=0 has a solution between x=0 and x=1
1 answer:
Answer:
Since the value of f(0) is negative and the value of f(1) is positive, then there is at least one value of x between 0 and 1 for which f(x) =0.
Step-by-step explanation:
The equation f(x) given is:
For x = 0. the value of the expression is:
For x = 1, the value of the expression is:
Since the value of f(0) is negative and the value of f(1) is positive, then there is at least one value of x between 0 and 1 for which f(x) =0.
In other words, there is at least one solution for the equation between x=0 and x=1.
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Answer:
Step-by-step explanation:
This is calculus, but I don't get fractions in the end. To maximize or minimize any function, you need to find the derivative of it, set it equal to 0, then solve for the critical values.
Our given equation is
x + y = 215 and we want to maximize the product, xy. Therefore,
y = 215 - x so its product in terms of x is
x(215-x) which is
. The derivative of this is
215 - 2x. Set it equal to 0 to maximize it.
215 - 2x = 0 so
-2x = -215 and
x = 107.5.
Sub this in to solve for y:
y + 107.5 = 215 and
y = 107.5
The product is 11556.25, not that you need it.