Answer on Question #60913 – Math – Calculus
Question
A) Solve the following simultaneous inequalities
8 x − 3 ≤ 1 4 [ x + 16 ] < x + 10 8x - 3 \leq \frac{1}{4} [x + 16] < x + 10 8 x − 3 ≤ 4 1 [ x + 16 ] < x + 10
B) Differentiate the following functions
y = x 3 + [ 6 x ] − 4 − [ 4 x ] ( 1 / 2 ) + 10 y = x^3 + \left[6x\right]^{-4} - \left[4x\right]^{(1/2) + 10} y = x 3 + [ 6 x ] − 4 − [ 4 x ] ( 1/2 ) + 10 y = ( [ 6 x ] − 3 + [ 4 x ] − 2 + 3 ) ( [ 2 x ] − 2 − [ 4 x ] − 4 + 5 ) y = \left(\left[6x\right]^{-3} + \left[4x\right]^{-2 + 3}\right) \left(\left[2x\right]^{-2 - \left[4x\right]^{-4 + 5}}\right) y = ( [ 6 x ] − 3 + [ 4 x ] − 2 + 3 ) ( [ 2 x ] − 2 − [ 4 x ] − 4 + 5 )
Solution
A)
Case 1. If [ x + 16 ] = x + 16 [x + 16] = x + 16 [ x + 16 ] = x + 16
8 x − 3 ≤ 1 4 [ x + 16 ] < x + 10 → 8x - 3 \leq \frac{1}{4} [x + 16] < x + 10 \rightarrow 8 x − 3 ≤ 4 1 [ x + 16 ] < x + 10 → 8 x − 3 ≤ x + 16 4 < x + 10 → 8x - 3 \leq \frac{x + 16}{4} < x + 10 \rightarrow 8 x − 3 ≤ 4 x + 16 < x + 10 → 32 x − 12 ≤ x + 16 < 4 x + 40 → 32x - 12 \leq x + 16 < 4x + 40 \rightarrow 32 x − 12 ≤ x + 16 < 4 x + 40 → { 32 x − 12 ≤ x + 16 , x + 16 < 4 x + 40 , \left\{
\begin{array}{l}
32x - 12 \leq x + 16, \\
x + 16 < 4x + 40,
\end{array}
\right. { 32 x − 12 ≤ x + 16 , x + 16 < 4 x + 40 , { 32 x − x ≤ 12 + 16 , 16 − 40 < 4 x − x , \left\{
\begin{array}{l}
32x - x \leq 12 + 16, \\
16 - 40 < 4x - x,
\end{array}
\right. { 32 x − x ≤ 12 + 16 , 16 − 40 < 4 x − x , { 31 x ≤ 28 , − 24 < 3 x , \left\{
\begin{array}{l}
31x \leq 28, \\
-24 < 3x,
\end{array}
\right. { 31 x ≤ 28 , − 24 < 3 x , { x ≤ 28 31 , x > − 8. \left\{
\begin{array}{l}
x \leq \frac{28}{31}, \\
x > -8.
\end{array}
\right. { x ≤ 31 28 , x > − 8.
Answer: ( − 8 , 28 31 ] (-8, \frac{28}{31}] ( − 8 , 31 28 ]
Case 2. If [ x + 16 ] = max { m ∈ Z ∣ m ≤ x + 16 } [x + 16] = \max\{m \in \mathbb{Z} \mid m \leq x + 16\} [ x + 16 ] = max { m ∈ Z ∣ m ≤ x + 16 }
8 x − 3 ≤ 1 4 [ x + 16 ] < x + 10 → 32 x − 12 ≤ [ x + 16 ] < 4 x + 40 → 8x - 3 \leq \frac{1}{4} [x + 16] < x + 10 \rightarrow 32x - 12 \leq [x + 16] < 4x + 40 \rightarrow 8 x − 3 ≤ 4 1 [ x + 16 ] < x + 10 → 32 x − 12 ≤ [ x + 16 ] < 4 x + 40 → → 32 x − 12 ≤ [ x ] + 16 < 4 x + 40 → \rightarrow 32x - 12 \leq [x] + 16 < 4x + 40 \rightarrow → 32 x − 12 ≤ [ x ] + 16 < 4 x + 40 → 32 x − 28 ≤ [ x ] < 4 x + 24 32x - 28 \leq [x] < 4x + 24 32 x − 28 ≤ [ x ] < 4 x + 24
Consider
[ x ] < 4 x + 24 ⇔ [ x < 4 x + 24 if 4 x + 24 is integer x < 4 x + 25 if 4 x + 24 is not integer ⇔ [ − 24 < 3 x if 4 x + 24 = k , k is integer − 25 < 3 x if 4 x + 24 ≠ k , k is not integer ⇔ [ x > − 8 if x = k 4 − 6 , k is integer x > − 25 3 if x ≠ k 4 − 6 , k is integer \begin{array}{l}
[x] < 4x + 24 \Leftrightarrow \left[ \begin{array}{l}
x < 4x + 24 \text{ if } 4x + 24 \text{ is integer} \\
x < 4x + 25 \text{ if } 4x + 24 \text{ is not integer}
\end{array} \right. \Leftrightarrow \\
\left[ \begin{array}{l}
-24 < 3x \text{ if } 4x + 24 = k, k \text{ is integer} \\
-25 < 3x \text{ if } 4x + 24 \neq k, k \text{ is not integer}
\end{array} \right. \Leftrightarrow \\
\left[ \begin{array}{l}
x > -8 \text{ if } x = \frac{k}{4} - 6, k \text{ is integer} \\
x > -\frac{25}{3} \text{ if } x \neq \frac{k}{4} - 6, k \text{ is integer}
\end{array} \right. \\
\end{array} [ x ] < 4 x + 24 ⇔ [ x < 4 x + 24 if 4 x + 24 is integer x < 4 x + 25 if 4 x + 24 is not integer ⇔ [ − 24 < 3 x if 4 x + 24 = k , k is integer − 25 < 3 x if 4 x + 24 = k , k is not integer ⇔ [ x > − 8 if x = 4 k − 6 , k is integer x > − 3 25 if x = 4 k − 6 , k is integer
If x = − 8 x = -8 x = − 8 8 ⋅ ( − 8 ) − 3 ≤ 1 4 [ − 8 + 16 ] < − 8 + 10 8 \cdot (-8) - 3 \leq \frac{1}{4} [-8 + 16] < -8 + 10 8 ⋅ ( − 8 ) − 3 ≤ 4 1 [ − 8 + 16 ] < − 8 + 10 x = − 8 x = -8 x = − 8 x = − 25 3 x = -\frac{25}{3} x = − 3 25 8 ⋅ ( − 25 3 ) − 3 ≤ 1 4 [ − 25 3 + 16 ] < − 25 3 + 10 8 \cdot \left(-\frac{25}{3}\right) - 3 \leq \frac{1}{4} \left[ -\frac{25}{3} + 16 \right] < -\frac{25}{3} + 10 8 ⋅ ( − 3 25 ) − 3 ≤ 4 1 [ − 3 25 + 16 ] < − 3 25 + 10 x = − 25 3 x = -\frac{25}{3} x = − 3 25 − 25 3 -\frac{25}{3} − 3 25 − 8 -8 − 8 x = − 81 10 x = -\frac{81}{10} x = − 10 81 8 ⋅ ( − 81 10 ) − 3 ≤ 1 4 [ − 81 10 + 16 ] < − 81 10 + 10 8 \cdot \left(-\frac{81}{10}\right) - 3 \leq \frac{1}{4} \left[ -\frac{81}{10} + 16 \right] < -\frac{81}{10} + 10 8 ⋅ ( − 10 81 ) − 3 ≤ 4 1 [ − 10 81 + 16 ] < − 10 81 + 10 − 64.8 ≤ 7 4 < 1.9 -64.8 \leq \frac{7}{4} < 1.9 − 64.8 ≤ 4 7 < 1.9
Consider
[ x ] ≥ 32 x − 28 ⇔ [ x ≥ 32 x − 28 if 32 x − 28 is integer x ≥ 32 x − 27 if 32 x − 28 is not integer ⇔ [ 31 x ≤ 28 if 32 x − 28 = m , m is integer ⇔ [ 31 x ≤ 27 if 32 x − 28 ≠ m , m is integer ⇔ \begin{array}{l}
[x] \geq 32x - 28 \Leftrightarrow \left[ \begin{array}{l}
x \geq 32x - 28 \text{ if } 32x - 28 \text{ is integer} \\
x \geq 32x - 27 \text{ if } 32x - 28 \text{ is not integer}
\end{array} \right. \Leftrightarrow \\
\left[ \begin{array}{l}
31x \leq 28 \text{ if } 32x - 28 = m, m \text{ is integer}
\end{array} \right. \Leftrightarrow \\
\left[ \begin{array}{l}
31x \leq 27 \text{ if } 32x - 28 \neq m, m \text{ is integer}
\end{array} \right. \Leftrightarrow \\
\end{array} [ x ] ≥ 32 x − 28 ⇔ [ x ≥ 32 x − 28 if 32 x − 28 is integer x ≥ 32 x − 27 if 32 x − 28 is not integer ⇔ [ 31 x ≤ 28 if 32 x − 28 = m , m is integer ⇔ [ 31 x ≤ 27 if 32 x − 28 = m , m is integer ⇔ [ x ≤ 28 31 if x = m 32 + 7 8 , m is integer x ≤ 27 31 if x ≠ m 32 + 7 8 , m is integer \left[ \begin{array}{l}
x \leq \frac{28}{31} \text{ if } x = \frac{m}{32} + \frac{7}{8}, m \text{ is integer} \\
x \leq \frac{27}{31} \text{ if } x \neq \frac{m}{32} + \frac{7}{8}, m \text{ is integer}
\end{array} \right. [ x ≤ 31 28 if x = 32 m + 8 7 , m is integer x ≤ 31 27 if x = 32 m + 8 7 , m is integer
If x = 28 31 x = \frac{28}{31} x = 31 28 8 ⋅ ( 28 31 ) − 3 ≤ 1 4 [ 28 31 + 16 ] < 28 31 + 10 8 \cdot \left(\frac{28}{31}\right) - 3 \leq \frac{1}{4} \left[\frac{28}{31} + 16\right] < \frac{28}{31} + 10 8 ⋅ ( 31 28 ) − 3 ≤ 4 1 [ 31 28 + 16 ] < 31 28 + 10 x = 28 31 x = \frac{28}{31} x = 31 28 x = 27 31 x = \frac{27}{31} x = 31 27 8 ⋅ ( 27 31 ) − 3 ≤ 1 4 [ 27 31 + 16 ] < 27 31 + 10 8 \cdot \left(\frac{27}{31}\right) - 3 \leq \frac{1}{4} \left[\frac{27}{31} + 16\right] < \frac{27}{31} + 10 8 ⋅ ( 31 27 ) − 3 ≤ 4 1 [ 31 27 + 16 ] < 31 27 + 10 x = 27 31 x = \frac{27}{31} x = 31 27 27 31 \frac{27}{31} 31 27 28 31 \frac{28}{31} 31 28 x = 9 10 x = \frac{9}{10} x = 10 9 8 ⋅ ( 9 10 ) − 3 ≤ 1 4 [ 9 10 + 16 ] < 9 10 + 10 8 \cdot \left(\frac{9}{10}\right) - 3 \leq \frac{1}{4} \left[\frac{9}{10} + 16\right] < \frac{9}{10} + 10 8 ⋅ ( 10 9 ) − 3 ≤ 4 1 [ 10 9 + 16 ] < 10 9 + 10
Answer: ( − 25 3 , − 8 ) ∪ ( − 8 , 27 31 ] \left(-\frac{25}{3}, -8\right) \cup \left(-8, \frac{27}{31}\right] ( − 3 25 , − 8 ) ∪ ( − 8 , 31 27 ]
B)
Case 1. Assuming [ 6 x ] = 6 x [6x] = 6x [ 6 x ] = 6 x [ 4 x ] = 4 x [4x] = 4x [ 4 x ] = 4 x [ 2 x ] = 2 x [2x] = 2x [ 2 x ] = 2 x
If
y = x 3 + ( 6 x ) − 4 − ( 4 x ) 1 2 + 10 , then y ′ = 3 x 2 − 4 1296 x − 5 − x − 1 2 = 3 x 2 − 1 324 x 5 − 1 x . \begin{array}{l}
y = x^{3} + (6x)^{-4} - (4x)^{\frac{1}{2}} + 10, \text{ then } \\
y' = 3x^{2} - \frac{4}{1296}x^{-5} - x^{-\frac{1}{2}} = 3x^{2} - \frac{1}{324x^{5}} - \frac{1}{\sqrt{x}}.
\end{array} y = x 3 + ( 6 x ) − 4 − ( 4 x ) 2 1 + 10 , then y ′ = 3 x 2 − 1296 4 x − 5 − x − 2 1 = 3 x 2 − 324 x 5 1 − x 1 .
If
y = ( ( 6 x ) 3 + ( 4 x ) 2 + 3 ) ( ( 2 x ) 2 − ( 4 x ) − 2 + 5 ) = = 864 x 5 + 64 x 4 + 1080 x 3 + 92 x 2 − 27 2 x − 3 16 x 2 + 14 , then y ′ = 4320 x 4 + 256 x 3 + 3240 x 2 + 184 x + 3 8 x 3 − 27 2 . \begin{array}{l}
y = ((6x)^{3} + (4x)^{2} + 3)((2x)^{2} - (4x)^{-2} + 5) = \\
= 864x^{5} + 64x^{4} + 1080x^{3} + 92x^{2} - \frac{27}{2}x - \frac{3}{16x^{2}} + 14, \text{ then } \\
y' = 4320x^{4} + 256x^{3} + 3240x^{2} + 184x + \frac{3}{8x^{3}} - \frac{27}{2}.
\end{array} y = (( 6 x ) 3 + ( 4 x ) 2 + 3 ) (( 2 x ) 2 − ( 4 x ) − 2 + 5 ) = = 864 x 5 + 64 x 4 + 1080 x 3 + 92 x 2 − 2 27 x − 16 x 2 3 + 14 , then y ′ = 4320 x 4 + 256 x 3 + 3240 x 2 + 184 x + 8 x 3 3 − 2 27 .
Case 2.
Assuming [ 6 x ] = max { m ∈ Z ∣ m ≤ 6 x } [6x] = \max\{m \in \mathbb{Z} \mid m \leq 6x\} [ 6 x ] = max { m ∈ Z ∣ m ≤ 6 x } [ 4 x ] = max { k ∈ Z ∣ k ≤ 4 x } [4x] = \max\{k \in \mathbb{Z} \mid k \leq 4x\} [ 4 x ] = max { k ∈ Z ∣ k ≤ 4 x }
[ 2 x ] = max { l ∈ Z ∣ l ≤ 2 x } . [2x] = \max\{l \in \mathbb{Z} \mid l \leq 2x\}. [ 2 x ] = max { l ∈ Z ∣ l ≤ 2 x } .
If
y = x 3 + ( [ 6 x ] ) − 4 − ( [ 4 x ] ) 1 2 + 10 , then y ′ = 3 x 2 , if x ≠ k 6 , x ≠ l 4 , k is integer, l is integer , y ′ does not exist if x = k 6 or x = l 4 , k is integer, l is integer . \begin{array}{l}
y = x^{3} + ([6x])^{-4} - ([4x])^{\frac{1}{2}} + 10, \text{ then } \\
y' = 3x^{2}, \text{ if } x \neq \frac{k}{6}, x \neq \frac{l}{4}, k \text{ is integer, } l \text{ is integer}, \\
y' \text{ does not exist if } x = \frac{k}{6} \text{ or } x = \frac{l}{4}, k \text{ is integer, } l \text{ is integer}.
\end{array} y = x 3 + ([ 6 x ] ) − 4 − ([ 4 x ] ) 2 1 + 10 , then y ′ = 3 x 2 , if x = 6 k , x = 4 l , k is integer, l is integer , y ′ does not exist if x = 6 k or x = 4 l , k is integer, l is integer .
If y = ( ( [ 6 x ] ) 3 + ( [ 4 x ] ) 2 + 3 ) ⋅ ( [ 2 x ] ) 2 − ( [ 4 x ] ) − 2 + 5 y = (([6x])^3 + ([4x])^2 + 3) \cdot ([2x])^2 - ([4x])^{-2} + 5 y = (([ 6 x ] ) 3 + ([ 4 x ] ) 2 + 3 ) ⋅ ([ 2 x ] ) 2 − ([ 4 x ] ) − 2 + 5
y ′ = 0 , if x ≠ k 6 , x ≠ l 4 , k is integer, l is integer ; y ′ does not exist if x = k 6 or x = l 4 , k is integer, l is integer . \begin{array}{l}
y' = 0, \text{ if } x \neq \frac{k}{6}, x \neq \frac{l}{4}, k \text{ is integer, } l \text{ is integer}; \\
y' \text{ does not exist if } x = \frac{k}{6} \text{ or } x = \frac{l}{4}, k \text{ is integer, } l \text{ is integer}.
\end{array} y ′ = 0 , if x = 6 k , x = 4 l , k is integer, l is integer ; y ′ does not exist if x = 6 k or x = 4 l , k is integer, l is integer .
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#339914 on Dec 2023