Math, asked by eegalapati, 1 month ago

the smaller circle touches the bigger circle and also its perpendicular diameters. if the radius of the smaller circle is 2cm , find the radius of bigger circle​

Answers

Answered by ᎷᎪᎠᎪᎡᎪ
11

Answer:

f′(x)f′(x) gives you the slope of ff in x

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0ln(1)=0ln(1)=0

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0ln(1)=0ln(1)=01<y⇔ln(y)>01<y⇔ln(y)>0

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0ln(1)=0ln(1)=01<y⇔ln(y)>01<y⇔ln(y)>0So we can write that

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0ln(1)=0ln(1)=01<y⇔ln(y)>01<y⇔ln(y)>0So we can write thatf′(x)>0⇔ln(x2x+1)>0⇔x2x+1>1f′(x)>0⇔ln(x2x+1)>0⇔x2x+1>1

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0ln(1)=0ln(1)=01<y⇔ln(y)>01<y⇔ln(y)>0So we can write thatf′(x)>0⇔ln(x2x+1)>0⇔x2x+1>1f′(x)>0⇔ln(x2x+1)>0⇔x2x+1>1f′(x)<0⇔ln(x2x+1)<0⇔x2x+1<1f′(x)<0⇔ln(x2x+1)<0⇔x2x+1<1

f′(x)f′(x) gives you the slope of ff in xQuite easily, if f′(x)f′(x) is positive, f(x)f(x) increases. If f′(x)f′(x) is negative, f(x)f(x) decreases.We know that, for y∈R∗+y∈R∗+0<y<1⇔ln(y)<00<y<1⇔ln(y)<0ln(1)=0ln(1)=01<y⇔ln(y)>01<y⇔ln(y)>0So we can write thatf′(x)>0⇔ln(x2x+1)>0⇔x2x+1>1f′(x)>0⇔ln(x2x+1)>0⇔x2x+1>1f′(x)<0⇔ln(x2x+1)<0⇔x2x+1<1f′(x)<0⇔ln(x2x+1)<0⇔x2x+1<1If x<−1

Attachments:
Similar questions