A function is increasing on an interval if for every point on that interval the first derivative is positive.

So we need to find the first derivative and then plug in the endpoints of our interval.

$\displaystyle h(x)=4x^3-5x^2-4x+7$

Find the first derivative by using the Power Rule $\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

$\displaystyle h'(x)=12x^2-10x-4$

Plug in the endpoints and evaluate the function.

$\displaystyle h'(2)=12\cdot2^2-10\cdot2-4=24$

$\displaystyle h'(3)=12\cdot3^2-10\cdot3-4=74$

Both are positive, so our function is increasing on the given interval.

The first step is to find the first derivative.

$\displaystyle y'=\frac{2x-2}{x^2-2x}$

Remember that the derivative of $\displaystyle ln(u) = \frac{u'}{u}$

Next, find the critical points, which are the points where $\displaystyle y'=0$ or undefined. To find the $\displaystyle 0$ points, set the numerator to $\displaystyle 0$, to find the undefined points, set the denomintor to $\displaystyle 0$. The critical points are $\displaystyle x=0,1,$ and $\displaystyle 2.$

The final step is to try points in all the regions $\displaystyle (-\infty,0) , (0,1), (1,2), (2,\infty)$ to see which range gives a positive value for $\displaystyle y'$.

If we plugin in a number from the first range, i.e $\displaystyle -1$, we get a negative number.

From the second range, $\displaystyle 0.5$, we get a positive number.

From the third range, $\displaystyle 1.5$, we get a negative number.

From the last range, $\displaystyle 3$, we get a positive number.

So the second and the last ranges are the ones where $\displaystyle y$ is increasing.

$\displaystyle f(x)$ is increasing when $\displaystyle f'(x)$ is positive (above the $\displaystyle x$-axis). This occurs on the intervals $\displaystyle (-\infty, -1), (0,2)$.

A function $\displaystyle f(x)$ is increasing if, for any $\displaystyle y>x$, $\displaystyle f(y)\geq f(x)$ (i.e the slope is always greater than or equal to zero)

Function E is the only function that has this property. Note that function E is increasing, but not strictly increasing

To find the increasing intervals of a given function, one must determine the intervals where the function has a positive first derivative. To find these intervals, first find the critical values, or the points at which the first derivative of the function is equal to zero.

For the given function, $\displaystyle f'(x)=3x^2-9$.

This derivative was found by using the power rule 

$\displaystyle \frac{d}{dx}x^n=nx^{n-1}$. 

When set equal to zero, $\displaystyle x=c=\pm\sqrt{3}$. Because we are only considering the open interval (0,5) for this function, we can ignore $\displaystyle c=-\sqrt{3}$. Next, we look the intervals around the critical value $\displaystyle c=\sqrt{3}$, which are $\displaystyle (0, \sqrt{3})$ and $\displaystyle (\sqrt{3}, 5)$. On the first interval, the first derivative of the function is negative (plugging in values gives us a negative number), which means that the function is decreasing on this interval. However for the second interval, the first derivative is positive, which indicates that the function is increasing on this interval $\displaystyle (\sqrt{3}, 5)$. 

Is f(x) increasing, decreasing, or flat at $\displaystyle x=-6$?

$\displaystyle f(x)=-3x^5+6x^3$

Recall that to find if a function is increasing or decreasing, we can use its first derivative. If f'(x) is positive, f(x) is increasing. If f'(x) is negative, f(x) is decreasing.

So, given:

$\displaystyle f(x)=-3x^5+6x^3$

We get

$\displaystyle f'(x)=-15x^4+18x^2$

Then:

$\displaystyle f'(x)=-15(-6)^4+18(-6)^2=-18792$

Therefore, f(x) is decreasing at the point, because f'(x) is negative.

Tell whether g(t) is increasing or decreasing on the interval [4,7]

$\displaystyle g(t)=4t^3-3t^2+6t$

To find increasing and decreasing, find where the first derivative is positive and negative. If g'(t) is positive, then g(t) is increasing and vice-versa.

$\displaystyle g(t)=4t^3-3t^2+6t$

$\displaystyle g'(t)=12t^2-6t+6$

Then,plug in the endpoints of [4,7] and see what you get for a sign.

$\displaystyle g'(4)=12(4)^2-6(4)+6=174$$\displaystyle g'(7)=12(7)^2-6(7)+6=552$

So, since g'(t) is positive on the interval, g(t) is increasing.

To find the interval(s) on which the function is increasing, we must find the intervals on which the first derivative of the function is positive. 

The first derivative of the function is:

$\displaystyle f'(x)=2x+1$

and was found using the rule

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now we must find the critical value, at which the first derivative is equal to zero:

$\displaystyle c=-\frac{1}{2}$

Now, we make the intervals on which we look at the sign of the first derivative:

$\displaystyle \left(-\infty , -\frac{1}{2}\right), \left(-\frac{1}{2}, \infty \right)$

On the first interval the first derivative is positive, while on the second it is negative. Thus, the first interval is our answer, because over this range of x values, the first derivative is positive and the function is increasing. 

If $\displaystyle f(x)$ is continuous everywhere and always increasing (i.e. $\displaystyle f'(x)>0$ for all $\displaystyle x$), then it must be true that after $\displaystyle f$ has attained its root, it can never do so again because it can't "return" to the $\displaystyle x$-axis. NOTE: this is not automatically true of functions that aren't continuous. As for the other choices, the possibility of at least having one more root is automatically false and a simple counterexample to the notion thay $\displaystyle f$ has to have an inflection point is a simple increasing lines. It has a constant positive derivative, but possesses no upward or downward concavity and has no inflection points.

Take the first derivative of $\displaystyle f(x)$:

$\displaystyle y'= \frac{6xlnx-3x}{ln^{2}x}$   by quotient rule

$\displaystyle f(x)$ is increasing whenever $\displaystyle y'$ is positive, that is, whenever both the numerator and denominator are of the same sign. The function $\displaystyle f(x)=lnx$ is certainly positive for all values of $\displaystyle x$ greater than $\displaystyle e$ because  and since 

$\displaystyle \frac{d}{dx}(lnx) = \frac{1}{x}$

is positive for all positive $\displaystyle x$, it is increasing on the interval, too. It will never be negative. For the same reason, the numerator is always positive. With the numerator and denominator always positive everywhere on the given interval, the derivative is always positive and the function is always increasing. So for any interval of nonzero length within $\displaystyle (e,\infty )$, $\displaystyle f(x)$ is increasing.

NOTE: Interestingly the opposite of the choice $\displaystyle f(x)$ > $\displaystyle 3x^{2}$ is also true. $\displaystyle 3x^{2}>lnx$ on the entire interval because at $\displaystyle x=e$, we have

$\displaystyle 3e^{2}>1=lne$. So the numerator is larger to begin with, and since:

$\displaystyle 6x > 1/x$  for all $\displaystyle x>e$ (or any $\displaystyle x>1$ for that matter), the derivative of the numerator is greater, too. This means the numerator will always be larger, so this condition coincides with the condition of $\displaystyle f'(x)$ being positive.