The given equation has a solution r in the interval 3<=r<=-2 Approximate the solution correct to two decimal places.x^(3)+2x^(2)-3x+7=0

Answer 1

The given equation x³+2x²-3x+7=0 has only one real solution which is about x=-3.45.

Plugging in x=-4 into x^3+2x^2-3x+7 gives the y value: y=(-4)^3+2(-4)^2-3(-4)+7=-13.

Plugging in x=-3 into x^3+2x^2-3x+7 gives the y value: y=(-3)^3+2(-3)^2-3(-3)+7=7.

So on the interval [-4,-3], the function goes from negative to positive. And since the function f(x)=x^3+2x^2-3x+7 is a polynomial, it is continuous on the interval [-4,-3] as well, and that means there has to be some x value where the function is zero.

Shrink the interval and continue this process where the function is positive on one end and negative on the other.

Eventually you get to the x interval [-3.455,-3.454]:

the y value at x=-3.455: y=f(-3.455)=-0.003371375, negative.

the y value at x=-3.454: y=f(-3.454)=0.015611336, positive.

Since the function goes from negative to positive on the interval [-3.455,-3.454], there has to be a root r that is strictly between -3.455 and -3.454. Rounding this to two decimal places gives x=-3.45.

Another way if you can graph is to graph the function f(x)=x^3+2x^2-3x+7 and find when it crosses the x-axis.

Yet another way is the Newton-Raphson Method of Root Finding from Calculus.