$(mx^3+3)*(2x^2+5x+2)-(8x^5+20x^4)=8x^3+6x^2+15x+6 \\ (mx^3*2x^2)+(mx^3*5x)+(mx^3*2)+(3*2x^2)+(3*5x)+(3*2) \\ -8x^5-20x^4=8x^3+6x^2+15x+6\\ \\ 2mx^5+5mx^4+2mx^3+6x^2+15x+6-8x^5-20x^4 \\ =8x^3+6x^2+15x+6 \\ \\ (2m-8) x^{5} +(5m-20) x^{4} +2mx^3+6x^2+15x+6 \\ =8x^3+6x^2+15x+6 \\ \\ (2m-8) x^{5} +(5m-20) x^{4} +(2m-8)x^3=0 \\ (m-4) 2x^{5} +(m-4) 5x^{4} +(m-4)2 x^3=0 \\ (m-4)( 2x^{5} + 5x^{4} +2 x^3)=0 \\ m=4$

6 0

3 years ago

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Suppose that f(5) = 1, f '(5) = 4, g(5) = −9, and g'(5) = 5. Find the following values.

stiv31 [10]

$\bf \begin{cases}
f(5)=1\qquad &f'(5)=4
\\\\
g(5)=-9\qquad &g'(x)=5
\end{cases}
\\\\

\begin{cases}
(f\cdot g)'(5)\impliedby \textit{product rule}
\\\\
\left( \frac{f}{g} \right)'(5)\impliedby \textit{quotient rule}
\end{cases}\\\\
-----------------------------\\\\
(f\cdot g)'(5)\implies f(x)\cdot g(x)\implies 
\begin{array}{llll}
f'(x)g(x)+f(x)g'(x)\\\\
x=5\\\\\
[4\cdot -9]+[1\cdot 5]
\end{array}
\\\\
-----------------------------\\\\
$

$\bf \left( \frac{f}{g} \right)'(5)\implies 
\begin{array}{llll}
\cfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\\\\
x=5\\\\
\cfrac{[4\cdot -9]-[1\cdot 5]}{(-9)^2}
\end{array}$

5 0

3 years ago