# Solve the equation. Answer the following questions: (a) Over what x - intervals is f increasing//decreasing? (b) Find all critical points of f and test each for local maximum and local minimum. (c) Over what x - intervals is f concave up//down? f(x) = 4 + 8x^{3} - x^{4}

Solve the equation. Answer the following questions:
(a) Over what x - intervals is f increasing//decreasing?
(b) Find all critical points of f and test each for local maximum and local minimum.
(c) Over what x - intervals is f concave up//down?
$f\left(x\right)=4+8{x}^{3}-{x}^{4}$
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Benedict
To find the critical points of the given function. Put first derivative of the function equal to zero.
$f\left(x\right)=4+8{x}^{3}-{x}^{4}$
${f}^{\prime }\left(x\right)=8\left(3{x}^{2}\right)-4{x}^{3}=0$
$24{x}^{2}-4{x}^{3}=0$
$4{x}^{2}\left(6-x\right)=0$
$x=0.6$
a) To find the increasing/decreasing of the function put any value of the interval in the first derivative of the function. If the value is positive (greater than zero) the function is increasing in the interval and if the value is negative (less than zero) the function is decreasing in the interval.Intervals: $\left(-\mathrm{\infty },0\right),\left(0,6\right),\left(6,\mathrm{\infty }\right)$
${f}^{\prime }\left(x\right)>0\in \left(-\mathrm{\infty },0\right)$
So, the function is increasing in interval $\left(-\mathrm{\infty },0\right).$
${f}^{\prime }\left(x\right)>0\in \left(0,6\right)$
So, the function is increasing in interval $\left(0,6\right).$
${f}^{\prime }\left(x\right)<0\in \left(6,\mathrm{\infty }\right)$
So, the function is decreasing in interval $\left(6,\mathrm{\infty }\right).$
b) To find the local maxima or local minima, put the critical points in the double derivative of the function. If the value is positive, then the point is local minima and if the value is negative, then the point is local maxima.
${f}^{\prime }\left(x\right)=24{x}^{2}-4{x}^{3}$
$f"\left(x\right)=48x-12{x}^{2}$
put $x=0,$
$f"\left(x\right)=0,so,x=0$ is the point of infection
put $x=6,$
$f"\left(x\right)=-144,so,x=6$ is the point of local maxima and maxima value is
$f\left(6\right)=4+8\left(6{\right)}^{3}-\left(6{\right)}^{4}=436$
c) To find whether the given function is concave up/ concave down, put any value in the double derivative from the interval to check. If the double derivative is positive the function is concave up and if the double derivative is negative the function is concave down.
$f"\left(x\right)=48x-12{x}^{2}$
for $\left(-\mathrm{\infty },0\right)$
$f"\left(x\right)=$ negative,
concave down at $\left(-\mathrm{\infty },0\right)$
for $\left(0,6\right)$
$f"\left(x\right)=$ positive,
concave up at $\left(0,6\right)$
for $\left(6,\mathrm{\infty }\right)$
$f"\left(x\right)=$ negative,
concave down at $\left(6,\mathrm{\infty }\right)$
a) increasing in $\left(-\mathrm{\infty },0\right),\left(0,6\right)$
decreasing in $\left(6,\mathrm{\infty }\right)$
b) critical points, $x=0.6$
local maxima, $x=6$
c) concave up in $\left(-\mathrm{\infty },0\right),\left(6,\mathrm{\infty }\right)$
concave up in $\left(0,6\right)$