# Thread: derivative function

1. ## derivative function

Consider the function ${f{{\left({x}\right)}}}={2}{{x}}^{{3}}+{9}{{x}}^{{2}}-{168}{x}+{3},\quad-{7}\leq{x}\leq{5}$.

This function has an absolute minimum value equal to

and an absolute maximum value equal to

2. ## Re: derivative function

what have you done with this problem?

3. ## Re: derivative function

Absolute max or min is where first derivative is zero.

$f(x) =& 2x^3 + 9x^2 -168x + 3$

\begin{align*} \frac{df}{dx} =& 6x^2 + 18x -168 = 0 \\ &x^2 + 3 x - 28 = 0.......\text{[after dividing both sides by } 6 \text{ ]}\\ &(x + 7)(x - 4) = 0......\text{[after factorization]}\end{align*}

So absolute max or min can be at $x = -7$ or $x = 4$. But to find out which one which you plug-in the values into the function and find out which one's larger than the other.

Now plug-in $x = -7$ and separately $x = 4$ into $f(x)$ you'll find that $f(-7) > f(4)$
The function has absolute max at $x = -7$

And has absolute min at $x = 4$

You can check it in the following graph to be sure:

4. ## Re: derivative function

Originally Posted by x3bnm
Absolute max or min is where first derivative is zero.

$f(x) =& 2x^3 + 9x^2 -168x + 3$

\begin{align*} \frac{df}{dx} =& 6x^2 + 18x -168 = 0 \\ &x^2 + 3 x - 28 = 0.......\text{[after dividing both sides by } 6 \text{ ]}\\ &(x + 7)(x - 4) = 0......\text{[after factorization]}\end{align*}

So absolute max or min can be at $x = -7$ or $x = 4$. But to find out which one which you plug-in the values into the function and find out which one's larger than the other.

Now plug-in $x = -7$ and separately $x = 4$ into $f(x)$ you'll find that $f(-7) > f(4)$
The function has absolute max at $x = -7$

And has absolute min at $x = 4$

You can check it in the following graph to be sure:

It should be pointed out that in general, absolute maxima and minima could lie at turning points (where the first derivative is 0) OR at endpoints of the function.

5. ## Re: derivative function

Originally Posted by Prove It
It should be pointed out that in general, absolute maxima and minima could lie at turning points (where the first derivative is 0) OR at endpoints of the function.
Good point Prove It. Thanks for that.