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for both of them i tried subbing in the x number and subbing it in the differemtiated version of both but none seem to work so i guess my method is completely wrong, please help

the answer for a should be 2/3 >0

answer for b should be 5/2 > 0

YEEEEEET Dec 17, 2018

#1**0 **

I do not know how you define "increasing or decreasing", but I did graph both of the functions to help you here.

I also think that the answers for (a) and (b) are supposed to be one of the words "increasing" or "decreasing". (hence, that is what the question asked for)

- PM

PartialMathematician Dec 17, 2018

#4**+2 **

\(f(x)=\frac{1}{3x^2}=\frac{1}{3}x^{-2}\\ f'(x) =\frac{-2}{3}x^{-3}\\ f'(-1) =\frac{-2}{3}\times (-1)^{-3}\\ f'(-1) =\frac{-2}{3*(-1)^{3}}\\ f'(-1)=\frac{-2}{3*-1}\\ f'(-1)=\frac{2}{3}\\ \)

Since the gradient at x=-1 is positive, the graph is increasing.

You can also see this from PM's graph. you can see that the gradient of the tangent to the curve is positive when x=-1

Melody Dec 17, 2018

#6**+2 **

\(f(x)=4\sqrt{x}+\sqrt[4]{x}\\ f(x)=4*x^{0.5}+x^{0.25}\\ f'(x)=2x^{-0.5}+0.25x^{-0.75}\\ f'(16)=2*16^{-0.5}+0.25*16^{-0.75}\\ f'(16)=2*4^{-1}+0.25*16^{3/4*-1}\\ f'(16)=2*4^{-1}+0.25*2^{3*-1}\\ f'(16)=2*4^{-1}+0.25*8^{-1}\\ f'(16)=\frac{2}{4}+\frac{0.25}{8}\\ f'(16)=\frac{16}{32}+\frac{1}{32}\\ f'(16)=\frac{17}{32}\\ \)

The gradient for part b is NOT 5/2

However, it is positive so the curve is increasing.

I really did not need to do all that. As soon as the curve was differentiated it was clear that the for all positive values of x the curve would be increasing. I mean the first derivative would be positive.

Melody Dec 17, 2018