Intermediate Value Theorem...2

[nycmathguy]

Calculus
Section 2.5

 

[MathLover1]

giiven: ln(x) =x - √x

Let f(x) = ln(x) - x + √x

f is continuous on the interval [2,3].

f(2) = ln(2) - 2 + √2 > 0

f(3) = ln(3) - 3 + √3 < 0

By the Intermediate value Theorem, there is at least one number, c, between 2 and 3 such that f(c) = 0.

So, ln(c) - c + √c = 0

Therefore, ln(c )= c + √c


as you stated above, ln(x)-x+sqrt(x)=0 solutions are

x = 1
x≈2.49091

as you can see on the graph, x≈2.49091 is between 2 and 3

 

[nycmathguy]

giiven: ln(x) =x - √x

Let f(x) = ln(x) - x + √x

f is continuous on the interval [2,3].

f(2) = ln(2) - 2 + √2 > 0

f(3) = ln(3) - 3 + √3 < 0

By the Intermediate value Theorem, there is at least one number, c, between 2 and 3 such that f(c) = 0.

So, ln(c) - c + √c = 0

Therefore, ln(c )= c + √c


as you stated above, ln(x)-x+sqrt(x)=0 solutions are

x = 1
x≈2.49091

as you can see on the graph, x≈2.49091 is between 2 and 3

I see my error. The x-value(s) found must lie between the given interval not between f(a) and f(b).