what is the magnitude of the force between a 25 mu c charge exerts on a -10 mu c charge 8.5cm aways. (step by step)
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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
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