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The is the statement of the Mean Value Theorem from your $te\times tb00k$ $Tneoren$ $m4.5$ $Ne8n$ Value Theorem Let f be continuous over $leclose$ interval $\left(a.b\right)aod$ $i$ $eo$ $xd$ $csnlc$ $\left(ab\right)$ $mmd$ exists at least one point cE $\left(a.b\right)$ $si1dhtba$ $f^{'}\left(c\right)=\dfrac {f\left(b\right)-f\left(a\right)} {b-a}$ What is the geometric interpretation of the conclusion of the theorem? O The tangent line to the graph of \(f\left(x\right)\) at $l\left(c1\right)$ is parallel to the secant line connecting \(\left(a,f\left(a\right)\right)\) and \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(f\left(lef\left(x\left(right\right)$ at \(c\) is the secant line connecting \(\left(a,f\left(a\right)\right)\) $ana$ \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph $of$ $\left(f\left(leF\left(x\left(right\right)\left(\right)at1\left(c1\right)is$ perpendicular to the secant line connecting \(\left(a,f\left(a\right)\right)\) and A(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(fleH\left(x\right)nght\right)\right)\right)$ $at$ $\left(c\right)\right)ishoizonta|$ $Tnere$ is more than one tangent line to the graph $0no$ $\left(flcf\left(x\right)night\right)\left($ $at1\left(c1\right)$

If the sum of two consecutive numbers is $45$ and one number is $X$ .This statement in the form of equation $1s:$ $\left(1$ Point) $\right)$ $○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ $○sx+1ef\left(x+2$ $r1gnt\right)=145s$ $sx+1x=45s$

$\right)c$ $c$ $co$ $3c3x$ $x$ Question 2 $1pts$ For each of the following questions, clearly select the ONE correct answer. Which of the following functions is increasing on the interval $\left(0,1\right)7$ $Ap$ $Ass$ Coe $-lnlef\left(x\right)ngh$ Col $x-ln\left(x\right)$ $x-1$ $Co$ $○y$ $xcos\left(2x\right)$ Diff Eva $O-\dfrac {3x} {x-1}-1trac\left(3x\right)\left(x-1\right)$ Fac Inte $○x$ $x$ $ln\left(x\right)$ Lim Seri Sim Solu $Ne\times t$ > Con $Con$ $Con$

$8$ $\left(1$ Point) $1\right)$ The\ reciprocal\\ $0+11\right)$ \left(\frac{2} $c\left(2\right)$ {5}\right)^0\ $\right)$ \ $1111s\right)$ $S$ $S1S$ $s3S$ $S4S$ $s2S$

Using the \emph{removal of first derivative} method, the differential equation \( \frac{d^{2}y} $\left(d\times n$ $\left(2\right)\right)+P|ffac\left(dy\right)\left(dx\right)+Qy=F$ $dx\right)+Qy=RN\right)$ is transformed as \). For, the differential equation \frac{d^{2}y} $\left(d^{n}\left(2\right)y\right)$ $dx$ $\left(2\right)+2x$ $\left(0C\left(dy\right)\left(dx\right)+\left(x$ $2+1\right)y=\times n3+3x\right)$ the value of $\left(11\right)$

In the following problem $a,b,$ b,and c represent REAL NUMBERS. The derivative of $g\left(x\right)=log _{a}\left(bx\right)+cx^{2}is$ $○$ $g^{1}\left(x\right)=\right)dfrac\left(b\right)log _{-}a\left(bx\right)\right)\left(N1n$ $a\right)+2cx1\right)$ $○$ $\left(g\left(x\right)=\right)dtracb$ $loga\left(bx\right)+2c\times \right)Nln$ a}\) $○$ $g^{'}\left(x\right)=\left(dfraC\left(a\right)log _{-}a\left(bx\right)\right)\left(ln$ $\right)+2c\times 1\right)\right)$ $○$ $g^{1}\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(ln$ $a1\right)$ $○$ $\left(\left(g^{1}\left(x\right)=b$ $log _{-}a\left(bx\right)+2cx1\right)$ $○$ $g\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(1ln$ $a|+2c1\right)$