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garik1379 [7]
4 years ago
8

Use the three conditions to show that function f(x)=2x^2 +3x - 1 is continuous at x=0.​

Mathematics
1 answer:
koban [17]4 years ago
4 0

Answer:

f(x)  = 2x^2 +3x - 1

f'(x) = 4x + 3

f''(x) = 4

f(0) = 2(0)^2 + 3(0) - 1 = -1 (valid)

f'(0) = 4(0) + 3 = 3 (valid)

f''(0) = 4 (valid)

=> f(x) is continous at x = 0

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Help me with question a please ! With full workings !
frosja888 [35]
A)


\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
Q&({{ 0}}\quad ,&{{ 2}})\quad 
%  (c,d)
P&({{ 0.5}}\quad ,&{{ 0}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}

\bf QP=\sqrt{(0.5-0)^2+(0-2)^2}\implies QP=\sqrt{0.5^2+2^2}
\\\\\\
QP=\sqrt{\left( \frac{1}{2} \right)^2+4}\implies QP=\sqrt{ \frac{1^2}{2^2}+4}\implies QP=\sqrt{\frac{1}{4}+4}
\\\\\\
QP=\sqrt{\frac{17}{4}}\implies QP=\cfrac{\sqrt{17}}{\sqrt{4}}\implies QP=\cfrac{\sqrt{17}}{2}

b)

since QR=QP, that means that QO is an angle bisector, and thus the segments it makes at the bottom of RO and OP, are also equal, thus RO=OP

thus, since the point P is 0.5 units away from the 0, point R is also 0.5 units away from 0 as well, however, is on the negative side, thus R (-0.5, 0)


c)

what's the equation of a line that passes through the points (-0.5, 0) and (0,2)?

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
Q&({{ 0}}\quad ,&{{ 2}})\quad 
%   (c,d)
R&({{ -0.5}}\quad ,&{{ 0}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{0-2}{-0.5-0}\implies \cfrac{-2}{-0.5}

\bf m=\cfrac{\frac{-2}{1}}{-\frac{1}{2}}\implies \cfrac{-2}{1}\cdot \cfrac{2}{-1}\implies 4
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-2=4(x-0)\implies y=4x+2\\
\left. \qquad   \right. \uparrow\\
\textit{point-slope form}
7 0
3 years ago
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