Enter the value of x! Thank you!

Answer:
Step-by-step explanation:
given:
f(x) = g(x)
f(x) = x³ - 3x² + 2
g(x) = x² - 6x + 11
find: the value of x
solution:
therefore, the value of x = 3
Step-by-step explanation:
[tex]
\underline{\bf{Given\::}}
Given:
\underline{\bf{To\:find\::}}
Tofind:
\underline{\bf{Explanation\::}}
Explanation:
\boxed{\bf{\frac{1}{f} =\frac{1}{v} -\frac{1}{u} }}}}
\begin{gathered}\longrightarrow\sf{\dfrac{1}{-10} =\dfrac{1}{v} -\dfrac{1}{-30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-10} +\dfrac{1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{-3+1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\cancel{\dfrac{-2}{30} }}\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-15} }\\\\\\\longrightarrow\sf{v=-15\:cm}\end{gathered}
⟶
−10
1
=
v
1
−
−30
1
⟶
v
1
=
−10
1
+
30
1
⟶
v
1
=
30
−3+1
⟶
v
1
=
30
−2
⟶
v
1
=
−15
1
⟶v=−15cm
\boxed{\bf{M \:A \:G \:N\: I \:F \:I \:C\: A\: T \:I \:O\: N :}}
MAGNIFICATION:
\begin{gathered}\mapsto\sf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance\:of\:image}{Distance\:of\:object} =\dfrac{v}{u} }\\\\\\\mapsto\sf{m=\cancel{\dfrac{-30}{-15}} }\\\\\\\mapsto\bf{m=2\:cm}\end{gathered}
↦m=
Heightofobject(O)
Heightofimage(I)
=
Distanceofobject
Distanceofimage
=
u
v
↦m=
−15
−30
↦m=2cm
Thus;
The magnification will be 2 cm .
[/tex]