Question

question 6
use your graphing calculator to solve the equation graphically for all real solutions
$x^3 - 5x^2 + 18 = 0$
solutions: $x = $
make sure your answers are accurate to at least two decimals
question help: video post to forum

Explanation:

Step1: Define the function

Let \( f(x) = x^3 - 5x^2 + 18 \). We need to find the real roots of \( f(x)=0 \) by graphing.

Step2: Analyze the graph

Using a graphing calculator, we plot the function \( y = x^3 - 5x^2 + 18 \). We look for the x - intercepts (where \( y = 0 \)) of the graph.
By analyzing the graph (either by using the zero - finding feature of the graphing calculator or by observing the behavior of the cubic function), we find that the function \( f(x)=x^3 - 5x^2 + 18 \) has one real root.
We can also use the rational root theorem to check for possible rational roots. The possible rational roots are factors of 18 divided by factors of 1, i.e., \( \pm1,\pm2,\pm3,\pm6,\pm9,\pm18 \).
- \( f(1)=1 - 5 + 18=14 eq0 \)
- \( f(2)=8 - 20 + 18 = 6 eq0 \)
- \( f(3)=27-45 + 18 = 0\)? No, \( 27-45 + 18=0\)? \( 27 + 18=45\), so \( 45-45 = 0\)? Wait, \( f(3)=3^3-5\times3^2 + 18=27 - 45+18 = 0\)? Wait, \( 27+18 = 45\), \( 45 - 45=0\). Wait, but if we consider the graph, maybe there was a miscalculation. Wait, no, let's recalculate: \( x = 3\): \( 3^3=27\), \( 5x^2=5\times9 = 45\), so \( 27-45 + 18=0\). Wait, but if we graph \( y=x^3-5x^2 + 18\), when \( x = 3\), \( y = 0\). But also, let's check the derivative \( f^\prime(x)=3x^2-10x=x(3x - 10)\). The critical points are at \( x = 0\) and \( x=\frac{10}{3}\approx3.33\). \( f(0)=18\), \( f(\frac{10}{3})=(\frac{10}{3})^3-5\times(\frac{10}{3})^2+18=\frac{1000}{27}-\frac{500}{9}+18=\frac{1000 - 1500}{27}+18=\frac{- 500}{27}+18=\frac{-500 + 486}{27}=\frac{-14}{27}\approx - 0.52\). So the function has a local maximum at \( x = 0\) (value 18) and a local minimum at \( x=\frac{10}{3}\) (value \(\approx - 0.52\)). So the graph crosses the x - axis once (since the local minimum is negative and the local maximum is positive, and as \( x ightarrow\infty\), \( f(x) ightarrow\infty\) and as \( x ightarrow-\infty\), \( f(x) ightarrow-\infty\)? Wait, no, for a cubic function \( ax^3+bx^2+cx + d\) with \( a>0\), as \( x ightarrow\infty\), \( y ightarrow\infty\) and as \( x ightarrow-\infty\), \( y ightarrow-\infty\). But our local maximum at \( x = 0\) is 18 (positive) and local minimum at \( x=\frac{10}{3}\) is \(\approx - 0.52\) (negative). So the graph crosses the x - axis three times? Wait, I made a mistake in the rational root calculation. Let's use the graphing calculator approach.
Using a graphing calculator (for example, on a TI - 84: enter \( Y1=x^3-5x^2 + 18\), then use the "zero" feature).
We find that the roots are:
- One real root and two complex roots? Wait, no, the discriminant of a cubic equation \( ax^3+bx^2+cx + d = 0\) is \( \Delta=18abcd - 4b^3d + b^2c^2-4ac^3 - 27a^2d^2\). For \( a = 1\), \( b=-5\), \( c = 0\), \( d = 18\), \( \Delta=18\times1\times(-5)\times0\times18-4\times(-5)^3\times18+(-5)^2\times0^2-4\times1\times0^3-27\times1^2\times18^2\)
\(=0 + 4\times125\times18+0 - 0-27\times324\)
\(=9000-8748 = 252>0\), so there are three distinct real roots.
Wait, my earlier derivative analysis was wrong. Let's recalculate the derivative: \( f^\prime(x)=3x^2-10x\). Setting to zero: \( 3x^2-10x = 0\Rightarrow x(3x - 10)=0\Rightarrow x = 0\) or \( x=\frac{10}{3}\approx3.33\).
\( f(0)=0^3-5\times0^2 + 18=18\)
\( f(\frac{10}{3})=(\frac{10}{3})^3-5\times(\frac{10}{3})^2+18=\frac{1000}{27}-\frac{500}{9}+18=\frac{1000 - 1500}{27}+18=\frac{-500}{27}+18=\frac{-500 + 486}{27}=\frac{-14}{27}\approx - 0.52\)
Now, let's check the values of the function at some points:
- For \( x=-2\): \( f(-2)=(-2)^3-5\times(-2)^2+18=-8 - 20 + 18=-10\)
- For \( x = 1\): \( f(1)=1-5 + 18=14\)
- For \( x = 2\): \( f(2)=8-20 + 18=6\)
- For \( x = 3\): \( f(3)=27-45 + 18=0\)
- For \( x = 4\): \( f(4)=64-80 + 18=2\)
- For \( x = 5\): \( f(5)=125-125 + 18=18\)

Wait, when \( x = 3\), \( f(x)=0\). But also, since \( f(-2)=-10\) and \( f(0)=18\), there is a root between \( x=-2\) and \( x = 0\). Let's use the Newton - Raphson method to find the roots.

Let's find the root between \( x=-2\) and \( x = 0\). Let \( x_0=-1\), \( f(-1)=-1-5 + 18=12\), \( f^\prime(-1)=3 + 10=13\). Newton - Raphson formula: \( x_{n + 1}=x_n-\frac{f(x_n)}{f^\prime(x_n)}\)
\( x_1=-1-\frac{12}{13}\approx-1 - 0.923=-1.923\)
\( f(-1.923)=(-1.923)^3-5\times(-1.923)^2+18\approx-7.13-5\times3.698 + 18\approx-7.13-18.49+18\approx-7.62\)
\( f^\prime(-1.923)=3\times(-1.923)^2-10\times(-1.923)\approx3\times3.698+19.23\approx10.994 + 19.23\approx30.224\)
\( x_2=-1.923-\frac{-7.62}{30.224}\approx-1.923 + 0.252=-1.671\)
\( f(-1.671)=(-1.671)^3-5\times(-1.671)^2+18\approx-4.66-5\times2.792+18\approx-4.66 - 13.96+18\approx-0.62\)
\( f^\prime(-1.671)=3\times(-1.671)^2-10\times(-1.671)\approx3\times2.792+16.71\approx8.376+16.71\approx25.086\)
\( x_3=-1.671-\frac{-0.62}{25.086}\approx-1.671+0.025=-1.646\)
\( f(-1.646)=(-1.646)^3-5\times(-1.646)^2+18\approx-4.45-5\times2.709+18\approx-4.45-13.545+18\approx0.005\)
So one root is approximately \( x\approx - 1.65\) (accurate to two decimals).

We know \( x = 3\) is a root (since \( f(3)=0\)). Now, let's factor the cubic: \( x^3-5x^2 + 18=(x - 3)(x^2-2x - 6)\) (using polynomial long division or synthetic division).
Now, solve \( x^2-2x - 6=0\) using the quadratic formula \( x=\frac{2\pm\sqrt{4 + 24}}{2}=\frac{2\pm\sqrt{28}}{2}=\frac{2\pm2\sqrt{7}}{2}=1\pm\sqrt{7}\)
\( \sqrt{7}\approx2.6458\), so \( 1+\sqrt{7}\approx3.6458\approx3.65\) and \( 1-\sqrt{7}\approx-1.6458\approx-1.65\)

So the real solutions are \( x\approx - 1.65\), \( x = 3\), and \( x\approx3.65\)

Answer:

\( x\approx - 1.65,3,3.65\) (accurate to two decimal places)