Question

graph $h(x) = x^4 - 3x^3 + x$. identify the $x$-intercepts and the points where the local maximums and local minimums occur. determine the intervals for which the function is increasing or decreasing. round to the nearest hundredth, if necessary.

from least to greatest, the $x$-intercepts are $x \approx \square$, $x = \square$, $x \approx \square$, and $x \approx \square$.

the local maximum is $(\square, \square)$.

the local minimums are $(\square, \square)$ and $(\square, \square)$.

the function is increasing when $x > \square$ and $\square < x < \square$.

the function is decreasing when $x < \square$ and $\square < x < \square$.

Explanation:

Step1: Find x-intercepts

To find the x-intercepts, we set \( h(x)=0 \), so \( x^{4}-3x^{3}+x = 0 \), which factors as \( x(x^{3}-3x^{2}+1)=0 \). One root is \( x = 0 \). For the cubic equation \( x^{3}-3x^{2}+1=0 \), we can use numerical methods (e.g., Newton-Raphson) or a graphing calculator. Let's approximate the roots:
- Let \( f(x)=x^{3}-3x^{2}+1 \). \( f(0)=1 \), \( f(1)=1 - 3+1=-1 \), \( f(2)=8 - 12 + 1=-3 \), \( f(3)=27-27 + 1 = 1 \). So there are roots in \( (0,1) \), \( (2,3) \), and another? Wait, \( f(-1)=-1 - 3+1=-3 \), \( f(0)=1 \), so also in \( (-1,0) \)? Wait, no, original function is \( x(x^{3}-3x^{2}+1) \), so the cubic \( x^{3}-3x^{2}+1 \) has three real roots? Wait, \( f(0)=1 \), \( f(1)=-1 \), \( f(2)=-3 \), \( f(3)=1 \). So using Newton-Raphson:
- For the root in \( (0,1) \): Let \( x_0 = 0.5 \), \( f(0.5)=0.125 - 0.75+1 = 0.375 \), \( f'(x)=3x^{2}-6x \), \( f'(0.5)=0.75 - 3=-2.25 \). \( x_1=0.5-\frac{0.375}{-2.25}\approx0.6667 \). \( f(0.6667)\approx(0.6667)^3-3(0.6667)^2 + 1\approx0.296 - 1.333+1\approx -0.037 \). \( f'(0.6667)=3(0.444)-6(0.6667)\approx1.333 - 4=-2.667 \). \( x_2=0.6667-\frac{-0.037}{-2.667}\approx0.653 \). So \( x\approx0.65 \).
- For the root in \( (2,3) \): Let \( x_0 = 2.5 \), \( f(2.5)=15.625 - 18.75+1=-2.125 \), \( f'(2.5)=3(6.25)-6(2.5)=18.75 - 15 = 3.75 \). \( x_1=2.5-\frac{-2.125}{3.75}\approx2.5 + 0.5667\approx3.0667 \). \( f(3.0667)\approx(3.0667)^3-3(3.0667)^2+1\approx28.7 - 28.2+1\approx1.5 \). \( x_2=3.0667-\frac{1.5}{3(3.0667)^2-6(3.0667)}\approx3.0667-\frac{1.5}{28.2 - 18.4}\approx3.0667 - 0.16\approx2.9067 \). \( f(2.9067)\approx(2.9067)^3-3(2.9067)^2+1\approx24.4 - 25.3+1\approx0.1 \). \( x_3=2.9067-\frac{0.1}{3(2.9067)^2-6(2.9067)}\approx2.9067-\frac{0.1}{25.3 - 17.4}\approx2.9067 - 0.012\approx2.8947 \approx 2.89 \).
- For the root in \( (-1,0) \): Let \( x_0=-0.5 \), \( f(-0.5)=-0.125 - 0.75+1 = 0.125 \), \( f'(-0.5)=3(0.25)-6(-0.5)=0.75 + 3 = 3.75 \). \( x_1=-0.5-\frac{0.125}{3.75}\approx-0.5 - 0.033\approx-0.533 \). \( f(-0.533)\approx(-0.533)^3-3(-0.533)^2+1\approx-0.152 - 0.849+1\approx0.0 \). So \( x\approx - 0.53 \).

So the x-intercepts are \( x\approx - 0.53 \), \( x = 0 \), \( x\approx0.65 \), \( x\approx2.89 \).

Step2: Find critical points (local max/min)

First, find the derivative \( h'(x)=4x^{3}-9x^{2}+1 \). Set \( h'(x)=0 \), so \( 4x^{3}-9x^{2}+1 = 0 \). We can use numerical methods or a graphing calculator to find the roots of this cubic equation. Let's approximate:
- Let \( g(x)=4x^{3}-9x^{2}+1 \). \( g(0)=1 \), \( g(1)=4 - 9+1=-4 \), \( g(2)=32 - 36+1=-3 \), \( g(3)=108 - 81+1 = 28 \), \( g(0.1)=0.04 - 0.09+1 = 0.95 \), \( g(0.2)=0.032 - 0.36+1 = 0.672 \), \( g(0.3)=0.108 - 0.81+1 = 0.298 \), \( g(0.4)=0.256 - 1.44+1=-0.184 \). So a root in \( (0.3,0.4) \). \( g(2)=-3 \), \( g(3)=28 \), so a root in \( (2,3) \). \( g(-1)=-4 - 9+1=-12 \), \( g(0)=1 \), so a root in \( (-1,0) \).

Using Newton-Raphson:
- For the root in \( (-1,0) \): Let \( x_0=-0.5 \), \( g(-0.5)=4(-0.125)-9(0.25)+1=-0.5 - 2.25+1=-1.75 \), \( g'(-0.5)=12(-0.5)^2-18(-0.5)=3 + 9 = 12 \). \( x_1=-0.5-\frac{-1.75}{12}\approx-0.5 + 0.1458\approx-0.3542 \). \( g(-0.3542)\approx4(-0.3542)^3-9(-0.3542)^2+1\approx4(-0.044)-9(0.125)+1\approx-0.176 - 1.125+1=-0.301 \). \( x_2=-0.3542-\frac{-0.301}{12(-0.3542)^2-18(-0.3542)}\approx-0.3542-\frac{-0.301}{1.51 - (-6.376)}\approx-0.3542 + 0.037\approx-0.3172 \approx - 0.32 \). \( g(-0.32)\approx4(-0.032768)-9(0.1024)+1\approx-0.131 - 0.9216+1=-0.0526 \). \( x_3=-0.32-\frac{-0.0526}{12(0.32)^2-18(0.32)}\approx-0.32-\frac{-0.0526}{1.2288 - (-5.76)}\approx-0.32 + 0.007\approx-0.313 \approx - 0.31 \).
- For the root in \( (0.3,0.4) \): Let \( x_0=0.35 \), \( g(0.35)=4(0.042875)-9(0.1225)+1=0.1715 - 1.1025+1=0.069 \), \( g(0.36)=4(0.046656)-9(0.1296)+1=0.1866 - 1.1664+1=0.0202 \), \( g(0.37)=4(0.050653)-9(0.1369)+1=0.2026 - 1.2321+1=-0.0295 \). So root in \( (0.36,0.37) \). Using linear approximation: \( x\approx0.36+\frac{0 - 0.0202}{-0.0295 - 0.0202}(0.37 - 0.36)\approx0.36 + 0.404\times0.01\approx0.364 \approx 0.36 \).
- For the root in \( (2,3) \): Let \( x_0=2.5 \), \( g(2.5)=4(15.625)-9(6.25)+1=62.5 - 56.25+1=7.25 \), \( g(2.2)=4(10.648)-9(4.84)+1=42.592 - 43.56+1=0.032 \), \( g(2.21)=4(10.793)-9(4.8841)+1=43.172 - 43.9569+1=0.2151 \)? Wait, no, \( g(2.2)=4(2.2)^3-9(2.2)^2+1=4(10.648)-9(4.84)+1=42.592 - 43.56+1=0.032 \), \( g(2.19)=4(2.19)^3-9(2.19)^2+1=4(10.4603)-9(4.7961)+1=41.8412 - 43.1649+1=-0.3237 \). So root in \( (2.19,2.2) \). \( x\approx2.2 \) (since \( g(2.2)\approx0.032 \) close to 0).

Now, we can use the second derivative test or analyze the sign changes of \( h'(x) \) to determine local max and min.

Step3: Classify critical points

- For \( x\approx - 0.31 \): Let's check the sign of \( h'(x) \) around it. For \( x < -0.31 \), say \( x=-1 \), \( h'(-1)=4(-1)^3-9(-1)^2+1=-4 - 9+1=-12 < 0 \). For \( -0.31 < x < 0.36 \), say \( x = 0 \), \( h'(0)=1 > 0 \). So \( x\approx - 0.31 \) is a local minimum (since derivative changes from negative to positive).
- For \( x\approx0.36 \): For \( -0.31 < x < 0.36 \), \( h'(x)>0 \); for \( 0.36 < x < 2.2 \), say \( x = 1 \), \( h'(1)=4 - 9+1=-4 < 0 \). So \( x\approx0.36 \) is a local maximum (derivative changes from positive to negative).
- For \( x\approx2.2 \): For \( 0.36 < x < 2.2 \), \( h'(x)<0 \); for \( x > 2.2 \), say \( x = 3 \), \( h'(3)=108 - 81+1=28 > 0 \). So \( x\approx2.2 \) is a local minimum (derivative changes from negative to positive).

Now, we can find the corresponding \( y \)-values for these critical points:
- For \( x\approx - 0.31 \): \( h(-0.31)\approx(-0.31)^4-3(-0.31)^3+(-0.31)\approx0.0092 - 3(-0.029791)-0.31\approx0.0092 + 0.0894 - 0.31\approx-0.2114 \approx - 0.21 \).
- For \( x\approx0.36 \): \( h(0.36)\approx(0.36)^4-3(0.36)^3+0.36\approx0.0168 - 3(0.046656)+0.36\approx0.0168 - 0.13997+0.36\approx0.2368 \approx 0.24 \).
- For \( x\approx2.2 \): \( h(2.2)\approx(2.2)^4-3(2.2)^3+2.2\approx23.4256 - 3(10.648)+2.2\approx23.4256 - 31.944+2.2\approx-6.3184 \approx - 6.32 \).

Step4: Determine increasing/decreasing intervals

- From the sign of \( h'(x) \):
- When \( x < -0.31 \), \( h'(x)<0 \), so function is decreasing.
- When \( -0.31 < x < 0.36 \), \( h'(x)>0 \), so function is increasing.
- When \( 0.36 < x < 2.2 \), \( h'(x)<0 \), so function is decreasing.
- When \( x > 2.2 \), \( h'(x)>0 \), so function is increasing.

Answer:

From least to greatest, the x-intercepts are \( x\approx - 0.53 \), \( x = 0 \), \( x\approx0.65 \), and \( x\approx2.89 \).

The local maximum is \( (0.36,0.24) \).

The local minimums are \( (-0.31,-0.21) \) and \( (2.2,-6.32) \).

The function is increasing when \( x > 2.2 \) and \( -0.31 < x < 0.36 \).

The function is decreasing when \( x < -0.31 \) and \( 0.36 < x < 2.2 \).

(Note: The values are approximate and rounded to the nearest hundredth. The exact values may vary slightly depending on the method of approximation.)