curve sketching help! (1 Viewer)

[z3192422]

give y= f(x)

sketch y=f'(x), y=1/(x), y=f(lxl), y=f(x+1)
plz explain in full detail>thnx

 

[hscishard]

y=f(x) is impossible to sketch
...
y=1/f(x)
x intercepts become vertical asymptotes
the vertical asymptotes become 0 in the reciprocal function
It should be easy from here. If it's low and positive, it'll be high and positive

y=f[|x|]
Pretty much two f(x) for x>0, with the y axis being the axis of symmetry

last one you just shift everything one space back

 

[Pwnage101]

y=1/f(x)
x intercepts become vertical asymptotes
the vertical asymptotes become 0 in the reciprocal function

Open circle at 0.

 

[iSplicer]

y=f(x) is impossible to sketch
...
y=1/f(x)
x intercepts become vertical asymptotes
the vertical asymptotes become 0 in the reciprocal function
It should be easy from here. If it's low and positive, it'll be high and positive

y=f[|x|]
Pretty much two f(x) for x>0, with the y axis being the axis of symmetry

last one you just shift everything one space back

For y=1/f(x)

dy/dx = -f'(x) /[f(x)]^2 (denominator is strictly positive, so its consideration regarding the sign of dy/dx is superfluous)

so we have: the sign of dy/dx = the opposite sign of f'(x). What does this tell us? When f(x) is decreasing, 1/f(x) will be increasing and vice versa. This should also be good help. You can differentiate it again to prove the concavity but for a 2 marker in the HSC, i'm not sure how useful this'll be.

 

[z3192422]

y=f(x) is impossible to sketch
sr, its y=f'(x) not f(x)

 

[z3192422]

thanks so much!!