# Thread: Differentiation

1. ## Differentiation

For the following function f(x), find f'(x) and any intervals in which f(x) is increasing:

$x^4 + 4x^3$
$f(x) = x^4 + 4x^3$
$f'(x) = 4x^3 + 12x^2$
$f'(x) = 0$
$4x^3+12x^2 = 0$
$4x^2(x+3)$

$4x^2$
$x^2 = \frac{1}{4}$
$x = \sqrt{\frac{1}{4}}$
$x = 0.5$ The answer is wrong. where did I go wrong?

$x+3$
$x = -3$

2. Originally Posted by looi76
For the following function f(x), find f'(x) and any intervals in which f(x) is increasing:
Here you should look at where it has a positive gradient.

3. Originally Posted by looi76
4x^2 = 0
x = 0 The answer is wrong. where did I go wrong?
There you go.

4. How is it possible? You mean I don't have to substitute $x$? Every time $x$ would be equal to $0$?

5. Originally Posted by looi76
$4x^3+12x^2 = 0$
$4x^2(x+3)$

$4x^2$
$x^2 = \frac{1}{4}$
$x = \sqrt{\frac{1}{4}}$
$x = 0.5$ The answer is wrong. where did I go wrong?

$x+3$
$x = -3$
$4x^3+12x^2 = 0$
$4x^2(x+3)$ <- Here you threw the "= 0" part away. Don't do that.

$4x^2(x+3) = 0$
$4x^2 = 0$ and $(x+3) = 0$

6. Originally Posted by janvdl
$4x^3+12x^2 = 0$
$4x^2(x+3)$ <- Here you threw the "= 0" part away. Don't do that.

$4x^2(x+3) = 0$
$4x^2 = 0$ and $(x+3) = 0$

for $(x+3) = 0$, you need to substitute to get $x = -3$
but why don't we substitute for $4x^2 = 0$ because if we did we would be getting $x = 0.5$

7. Originally Posted by looi76
for $(x+3) = 0$, you need to substitute to get $x = -3$
but why don't we substitute for $4x^2 = 0$ because if we did we would be getting $x = 0.5$
$4x^2 = 0$

$x^2 = \frac{0}{4}$

$x^2 = 0$

$x = \sqrt{0}$

$x = 0$

8. You have two points: x=-3, and x=0.
Test the areas around each point. I chose x=-4, x=-1, and x=1

$f\prime(-4) = 4(-4)^3+12(-4)^2 = -64$
$f\prime(-1) = 4(-1)^3+12(-1)^2 = 8$
$f\prime(1) = 4(1)^3+12(1)^2 = 16$

So we know that before x=-3, the derivative of our function is less than zero, which means that our function is decreasing at this time.
Between x=-3 and x=0 the derivative becomes positive, so our function is increasing.
After x= 0 our function is still positive so our function is still increasing.

This means that our function is increasing at all points after x=-3... EXCEPT that we know the derivative equals zero at the point x=0, so at this point, it is neither increasing nor decreasing.

So f(x) is increasing on $[-3, 0)\cup(0, \infty)$

Originally Posted by looi76
for $(x+3) = 0$, you need to substitute to get $x = -3$
but why don't we substitute for $4x^2 = 0$ because if we did we would be getting $x = 0.5$
No, you would get x=0 as janvdl showed

You mistakenly set 4x^2 equal to 1, which is where you got 1/4, but you should have kept it equal to zero (because if it equals zero, then it is irrelevant what x+3 equals, since it will be multiplied by zero), and 0/4 = 0, and the square root of 0 = 0 so x=0

9. Originally Posted by looi76
For the following function f(x), find f'(x) and any intervals in which f(x) is increasing:

$x^4 + 4x^3$
$f(x) = x^4 + 4x^3$
$f'(x) = 4x^3 + 12x^2$
$f'(x) = 0$
$4x^3+12x^2 = 0$
$4x^2(x+3)$

$4x^2$
$x^2 = \frac{1}{4}$
$x = \sqrt{\frac{1}{4}}$
$x = 0.5$ The answer is wrong. where did I go wrong?

$x+3$
$x = -3$
y = f(x) increasing => $f'(x) > 0$.

Solving non-linear inequalities is always easiest and safest using a graph. So ....

Draw a graph of $f'(x) = 4x^3 + 12x^2 = 4x^2 (x + 3)$. You want the values of x for which this graph is ABOVE the x-axis.