X^3+5x-4=0 Show this equation has a solution between x=0 and x=1.
(The question is two marks)
I found out how to answer the question myself but for others wondering how, I solved it by subbing in the numbers and stating there is a change of sign. This would ensure the full two marks.
F(x)=x^3+5x-4
F(0)=(0)^3+5(0)-4
F(0)= -4
F(1)=(1)^3+5(1)-4
F(1)=2
There's a change of sign therefore the solution will lie between 0 and 1.
Exactly, we're looking for the change of sign which implies the presence of a root. It may be intuitive, but this fact is dependent on the Intermediate Value theorem, as stated above.
f(x)=x^3+5x-4
f(0)=-4
f(1)=2
Since f(x) is continuous (as are all polynomials), there exists a value c such that f(c)=0 (-4≤0≤2, and 0≤c≤1) by the Intermediate Value Theorem.
Well, solving an equation is like trying to find a hidden treasure. And in this case, we're looking for a treasure between x=0 and x=1.
Now, let me tell you a little secret. Like a mischievous clown, I like to play pranks on equations. So, here's what I'll do. I'll substitute both x=0 and x=1 into the equation and see what happens.
When we substitute x=0, we get 0^3 + 5(0) - 4 = -4. Uh-oh, looks like we missed the treasure!
But don't worry, we're not giving up yet! Let's try x=1. Plugging in x=1 gives us 1^3 + 5(1) - 4 = 2. Hurray! We struck gold!
So, my mathematically adventurous friend, we've shown that the equation indeed has a solution between x=0 and x=1. And we did it with a little mathematical clowning around. Good luck on your treasure hunt!
To show that the equation has a solution between x=0 and x=1, we can use the Intermediate Value Theorem.
The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b], and if f(a) and f(b) have different signs (meaning that f(a) < 0 and f(b) > 0, or vice versa), then the function must have at least one root (or zero) between a and b.
In this case, we have the equation f(x) = x^3 + 5x - 4 = 0.
First, let's evaluate f(0):
f(0) = 0^3 + 5(0) - 4 = -4.
Next, let's evaluate f(1):
f(1) = 1^3 + 5(1) - 4 = 2.
Since f(0) < 0 and f(1) > 0, we have f(0) < 0 < f(1), which implies that f(x) changes sign between x = 0 and x = 1.
Therefore, by the Intermediate Value Theorem, the equation x^3 + 5x - 4 = 0 must have at least one solution between x = 0 and x = 1.