Home > Calculus > Interpreting the Sign of the First Derivative (Increasing and Decreasing Functions)

Is #f(x)=1/x-1/x^3+1/x^5# increasing or decreasing at #x=2#?

Answer 1

Paisley Johnson

Decreasing.

#f'(x) = -1/x^2 + 3/x^4 - 5/x^6#

#f'(2) = -9/64#

Decreasing at #x = 2#.

To see why, consider a small change in #x#, #deltax#, near the neighborhood of #x=2#.

We can approximate #f(2+deltax)# as

#f(2) + f'(2)deltax#.

This approximation is most accurate for small values of #deltax#.

Since #f'(2)# is negative,

for sufficiently small values of #deltax > 0#, #f(2+deltax)

and for sufficiently small values of #deltax < 0#, #f(2+deltax)>f(2)#.

Therefore, #f(x)# is decreasing at #x=2#.

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Answer 2

Charles Anctil

To determine whether $f(x) = \frac{1}{x} - \frac{1}{x^3} + \frac{1}{x^5}$ is increasing or decreasing at $x = 2$ , we need to evaluate its derivative at that point. The derivative of $f(x)$ with respect to $x$ is $f'(x)$ . After finding $f'(x)$ , we substitute $x = 2$ into the derivative function.

Taking the derivative of $f(x)$ with respect to $x$ , we get: $f'(x) = -\frac{1}{x^2} + \frac{3}{x^4} - \frac{5}{x^6}$

Substituting $x = 2$ into $f'(x)$ : $f'(2) = -\frac{1}{2^2} + \frac{3}{2^4} - \frac{5}{2^6}$ $f'(2) = -\frac{1}{4} + \frac{3}{16} - \frac{5}{64}$ $f'(2) = -\frac{16}{64} + \frac{12}{64} - \frac{5}{64}$ $f'(2) = -\frac{9}{64}$

Since $f'(2)$ is negative ( $-\frac{9}{64}$ ), it indicates that $f(x)$ is decreasing at $x = 2$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.