Given f'(x) = (4x^2+5x+1)/(x-8)^(1/3) sketch f(x) where f(x) is a continuous function we are given the function (4x^2+5x+1)/(x-8)^(1/3) which is the first derivative of some continuous function...
Given f'(x) = (4x^2+5x+1)/(x-8)^(1/3) sketch f(x) where f(x) is a continuous function
we are given the function (4x^2+5x+1)/(x-8)^(1/3) which is the first derivative of some continuous function f(x).
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Let the derivative function be called `f'(x)`
First of all, notice that there is a singularity at `x=8` because here we are dividing by zero.
Next, factorise the top of the function giving
`f'(x) = (4x^2 + 5x+1)/(x-8)^(1/3) = ((4x+1)(x+1))/(x-8)^(1/3)`
This tells us that `f(x)` has turning points/points of inflection at `x=-1/4`, `x=-1` because the gradient of `f(x)`, `f'(x)`, is zero there.
Examine these points to determine whether they are maxima/minima/turning points.
To do this we examine the sign of the second derivative `f''(x) = d/(dx) f'(x)`
Now `f''(x) = d/(dx) (4x^2 + 5x+1)/(x-8)^(1/3) = (8x+5)/(x-8)^(1/3) -(1/3)(1)(4x^2+5x+1)/(x-8)^(-2/3)`
`= (8x+5)/(x-8)^(1/3) -(1/3)((4x+1)(x+1))/(x-8)^(-2/3)`
At `x=-1/4`, ` ``f''(x) approx (5-2)/(-8)^(1/3) = -3/2` which is negative implying there is a maximum at `x=-1/4`
At `x=-1`, `f''(x) approx (5-8)/(-8)^(1/3) = 3/2` which is positive implying there is a minimum at `x=-1`
Now look at what is happening around the value `x=8`
When `x=7`, `f'(x) = (4(49)+5(7)+1)/(-1)^(1/3)` which is negative
When `x=9`, `f'(x) = (4(81)+5(9)+1)/1^(1/3)` which is positive
Therefore as `x-> 8` from below, `f(x) -> -oo` and as `x-> 8` from above, `f(x) -> oo`
There is a minimum at `x=-1`, a maximum at `x=-1/4`
The gradient is negative over `(-oo,-1)`, positive over `(-1,-1/4)`, negative over `(-1/4,8)` and positive over `(8,oo)`
A graph` ` `g(x)` that has these properties is
`g(x) = (4/3x^3 + 5/2x^2 +x)/(x-8)` , `x < 8`
`= -1/(x-8)` , `x>8`
Minimum at -1, maximum at -1/4
Decreasing over (-inf,-1), increasing over (-1,-1/4),decreasing over (-1/4,8), increasing over (8,inf)
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