Question

please please help I cannot solve this​

Answer 1

Answer:

A function can be said to be increasing or decreasing based on it's First derivative.

If the first derivative f'(x) is greater than zero, then the given function is increasing. If f'(x) is less than zero, then the given function is decreasing.

According to the question, the given function is:

f(x) = 2x-1 / 3x + 4

Let us consider the numerator as 'u' and denominator as 'v'. Now the given function is of the form: u/v. Hence the differentiation of (u/v) is given as:

→ d(u/v) / dx = vdu - udv / v²

So, Differentiating the numerator with respect to 'x' we get:

→ d ( 2x -1 ) / dx

→ du = 2

Similarly differentiating the denominator with respect to 'x' we get:

→ d ( 3x + 4 ) / dx

→ dv = 3

So Differentiating the function we get:

→ d ( f(x) ) / dx = vdu - udv / v²

→ d ( f(x) ) / dx = (3x + 4 )(2) - (2x - 1 )(3) / (3x + 4)²

→ f'(x) = [ 6x + 8 - 6x + 3 ] / ( 3x + 4 )²

→ f'(x) = ( 11 ) / ( 3x + 4 )²

Now we know that, any number which is squared will always be greater than zero. Similarly the numerator 11 is also greater than zero.

Hence f'(x) > 0

Therefore the given function is increasing for all real values of 'x'.

Hence Proved!!