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3 years ago

If f(x) = x^2 + 1 and g(x) = 3x + 1, find 2f(1) + 3g(4)

2 answers:

8 0

$\bf \begin{cases}
f(x) = x^2 + 1\\\\
g(x) = 3x + 1
\end{cases}\quad 
\begin{cases}
f(1)=\boxed{1}^2+1\to &\boxed{?}\\\\
g(4)=3(\boxed{4})+1\to &\boxed{?}
\end{cases}
\\\\\\
2f(1)=2\cdot \boxed{?}\qquad \qquad 3g(4)=3\cdot \boxed{?}$

weeeeeb [17]3 years ago

3 0

2f(1)+3g(4)=43

2*(1^2+1)=4
3*(3(4)+1)=39

4+39=43

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Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

The positive value of $k$ will result in exactly one real root is approximately 0.028.

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