GRAPH DAY

Answer to #31

f"(x)=0, when 2x=log(-x)

Since log is gefined only for +ve values -x must be +ve, so x must be -ve.

When x is -ve, LHS is -ve. Log of a number is -ve only when the number is a fraction, so x is a fraction.

Therefore f(x) has the least value at a certain -ve fractional value of x.

Clearly function goes to -∞ on the -ve side (limiting x to -∞)

And to ∞ on +ve side (limiting x to ∞)

So the graph of f(x) will be as depicted

[339]

please check this one and correct me if wrong

my first try to add 2 functions

1 more good graph[1]

DRAW :
e^2x+x²

if u wanna try the rest then try them
or i will do the questions

ha ha ha ha ha ha ha ha

good work KALYAN

now there r some more
now some interesting 1's

draw the following curves as much as possible

1)sin1/x
2)xsin1/x
3)x²sin1/x
4)x e^1/x
5)x⁶-3x⁴+3x²-5
6)1/2sin2x +cosx
7)x+ln(x²-1)

yahan pe main kya pagalon ki tarah questions hi post karta jaoon

tell me if u wanna me to close this thread

if there would be no more replies then 4 me this thread will die[16]

sir jaise aap chahein

mera show to flop show nikla [17][2]

shayad aapke saath kuch hit ho jaye

tank u sir jee

2)

subash now repair ur mistake by giving the answers to #6 and #9[1]

y kabhi -ve nahi huunda!!

SOME MORE QUESTIONS

3)log(cosec x)

4)log(x²+1/2x) + sec^-1x

5)sec^-1(tan^-1cosx)

coming to the second graph

question becomes

2^-x+sin2x

which can be done by addition of graphs i think

Remember {x} ranges between [0,1)

graph would hav still been wrong if it were -1

The graph as drawn by Subash (as in post #4) iis wrong. You should note that the function
f(x)=2^{\{x\}} is periodic. The graph will look like that drawn by kalyan in post #12.