Just in general. I was just wanting some steps on how I should go about sketching f(x)...
I know that Integration is the area of the curves...
Thanks!
-qbkr21
we could use integration, but we don't have to
this is similar to those "fully analyse the curve" questions, except we go a bit backwards.
given f ' (x) and f '' (x) we can find:
max and mins
concavity
intervals of increase and decrease
with that information we can tell exactly what f(x) will look like
in general, we know that a graph of x^n where smaller powers of x are accounted for will have n-1 turns and n roots.
so a graph of y = ax^5 + bx^4 + cx^3 + dx^2 + ex + d will in general, have 5 roots and 4 turns
now the derivative cuts down one power, so for a general x^5 graph, f ' (x) would look like a general x^4 and f '' (x) would look like an x^3 graph.
the above example looks like an x^3 graph, so we expect f ' (x) to look like an x^4 graph and f(x) to look like an x^5 graph. we can get the specific graph by doing some analysis, but you said in general, so let's not do that now...unless you want to. it would be a pain, since such questions never ask you for specific graphs, i dont think
if you are inclined to examine this question further, i can start you off by saying this. the graph of f '' (x) cuts the x-axis at -2, 2 and 4. so these are the points where the second derivative is zero, and thus these are the possible inflection points. that's the first peice of important info. we can construct the formula for this graph without even using calculus. since it's a general x^3 graph with roots -2, 2 and 4, we know it will be the graph where
y = (x + 2)(x - 2)(x - 4)
expand that and you have the graph of f '' (x).
well, those probably aren't completely accurate. remember when intergrating we obtain an aribitrary constant.
so f ' (x) = (1/4)x^4 - (4/3)x^3 - 2x^2 + 16x + C
and f(x) = (1/20)x^5 - (1/3)x^4 - (2/3)x^3 + 8x^2 + Cx + D
your functions will only be accurate if constants C and D are zero, and innaccurate otherwise