Finding the x-value of Max, Min, & Inflexion Point

[markosheehan]

View attachment 6299

i know how to do the first two parts the x value of the max is 0 and of the minimum is 8 and 4 is the x value of the inflexion point. however i don't know how to do part 3

 

[MarkFL]

Well, for part i) this can be explained using the fact that the first and second derivatives to not share a common root. I agree with your assessment of the $x$-values for the local extrema and point of inflection.

To answer part iii), we need to begin with:

\(\displaystyle f(x)=ax^3-bx^2\)

To find the critical values, we need to compute $f;(x)$ and equate this to zero and solve for $x$:

\(\displaystyle f'(x)=4ax^2-2bx=2x(2ax-b)=0\)

Hence:

\(\displaystyle x\in\left\{0,\frac{b}{2a}\right\}\)

From the graph, we know that the $x$ value of the local minimum must be:

\(\displaystyle x=\frac{b}{2a}\)

For the point of inflection, we set $f''(x)=0$:

\(\displaystyle f''(x)=8ax-2b=2(4a-b)=0\)

Hence, the $x$-value of the point of inflection must be:

\(\displaystyle x=\frac{b}{4a}\)

And now we have enough information to answer part iv). :D