Hello Natnat,
We are given the function:
$$f(x)=-5x^4+4x$$
a) Determine the x and y intercepts
To find the $x$-intercept(s), we equate $y=f(x)$ to zero, and solve for $x$:
$$f(x)=-5x^4+4x=-x\left(5x^3-4 \right)=0$$
Equating each factor to zero (applying the zero-factor property, we find:
$$x=0$$
$$5x^3-4=0\implies x=\sqrt[3]{\frac{4}{5}}$$
Hence, the $x$-intercepts are:
$$(0,0),\,\left(\sqrt[3]{\frac{4}{5}},0 \right)$$
To find the $y$-intercept, we equate $x$ to zero, and solve for $y$:
$$y=-5(0)^4+4(0)=0$$
Thus, the $y$-intercept is:
$(0,0)$
b) Determine the coordinates of the critical points
To find the critical values, we need to equate the first derivative of the function to zero:
$$f'(x)=-20x^3+4=-4\left(5x^3-1 \right)=0$$
Thus the critical value is:
$$x=\sqrt[3]{\frac{1}{5}}=\frac{1}{\sqrt[3]{5}}$$
And so the coordinates of the critical point is:
$$\left(\frac{1}{\sqrt[3]{5}},f\left(\frac{1}{\sqrt[3]{5}} \right) \right)=\left(\frac{1}{\sqrt[3]{5}},\frac{3}{\sqrt[3]{5}} \right)$$
c) Identify the nature of critical points
Using the second derivative test (since we will need the second derivative later to discuss concavity anyway), we first need to compute the second derivative:
$$f''(x)=-60x^2$$
We can see that for any non-zero value of $x$, the second derivative is negative, and so the critical point we found in part b) must be a maximum, and since there is only one turning or critical point, we may conclude that this critical point is the global maximum.
d) Determine the coordinates of the points of inflection
Equating the second derivative to zero, we find:
$$f''(x)=-60x^2=0$$
We find that $x$ is a root of even multiplicity, and so the second derivative will not change sign across this repeated root, hence we conclude that this function has no inflection points.
e) Graph the function
Let's wait until the end to do this once we have completed the analysis of the function.
f) State the intervals of increase and decrease
From part b) we found the critical value:
$$x=\frac{1}{\sqrt[3]{5}}$$
This one critical point divides the domain of all reals into two sub-intervals. Because we have already determined the extremum assocaited with this critical value is the global maximum, we may conclude:
$$\left(-\infty,\frac{1}{\sqrt[3]{5}} \right)$$ $f(x)$ is increasing.
$$\left(\frac{1}{\sqrt[3]{5}},\infty \right)$$ $f(x)$ is decreasing.
g) State the intervals of concavity
We have already determined that the function's second derivative is negative everywhere except for $x=0$. Hence the function is concave down on:
$$(-\infty,0)\,\cup\,(0,\infty)$$
Okay, now let turn our attention to sketching the graph of the function.
From part a) we found the intercepts, for a total of 2 points to plot:
$$(0,0),\,\left(\sqrt[3]{\frac{4}{5}},0 \right)\approx(0,0),(0.93,0)$$
From part b) we found the critical point:
$$\left(\frac{1}{\sqrt[3]{5}},\frac{3}{\sqrt[3]{5}} \right)\approx(0.58,1.75)$$
Now we know the function is increasing everywhere to the left of this critical and decreasing everywhere to the right of it. We also know the function is concave down everywhere except at the origin. Hence, putting all of this together, we obtain a graph closely resembling the following function plot:
View attachment 1843
