Sketch the graph of the given function. Use technology to approximate the interc

Question

Sketch the graph of the given function. Use technology to approximate the intercepts, coordinates of extrema, and points of inflection to one decimal place. Check your sketch using technology. HINT [See Example 1.]

g(t) = 1/4 t^4 ? 2/3 t^3 + 1/2 t^2

(a) Indicate the t- and y-intercepts. (If an answer does not exist, enter DNE.)

(t, y) =

(t, y) =

(b) Indicate any extrema. (If an answer does not exist, enter DNE.)

(t, y) =

(c) Indicate any points of inflection. (Order your answers from smallest to largest t. If an answer does not exist, enter DNE.)

(t, y) =

(t, y) =

(d) Indicate the behavior near singular points of f.

y ? +? as t ? 0

y ? 0 as t ? 0

y ? +? as t ? 1

y ? ?? as t ? ±1

The function is defined everywhere on the domain.

(e) Indicate the behavior at infinity.

y ? +? as t ? ±?

y ? ?? as t ? ??; y ? +? as t ? +?

y ? ?? as t ? ±?

y ? +? as t ? ??; y ? ?? as t ? +?

The domain of the function does not extend to infinity.

Please show work. Thank you!

(t, y) =

(t, y) =

Explanation / Answer

t-int :
t^4/4 - 2t^3/3 + t^2/2 = 0

Mult all over by 12 :
3t^4 - 8t^3 + 6t^2 = 0

t^2(3t^2 - 8t + 6) = 0

3t^2 - 8t + 6 = 0 ---> no solution

So, t-int is (0,0)

y-int :
Plug t = 0 , we get y = 0

So, (0,0)

--------------------------------------------------------------

Extrema :
Deriving :
g' = t^3 - 2t^2 + t = 0
t(t^2 - 2t + 1) = 0
t(t - 1)(t - 1) = 0
t = 0 and t = 1

g'' = 3t^2 - 4t + 1

g''(0) = 1
So, t = 0 is a MINIMUM

g''(1) = 0
So, t = 1 is NEITHER A MIN NOR A MAX

When t = 0, y = 0

So, (0,0) is a relative min -----> ANS

--------------------------------------------------------------

c)
Inf pts :
Deriivng :
g'' = 3t^2 - 4t + 1 = 0

(3t - 1)(t - 1) = 0

t = 1 and t = 1/3

When t = 1, we get y = 1/12
When t = 1/3, we get y = 11/324

So,
inf pts are
(1 , 1/12)
(1/3 ,11/324) -----> ANS

------------------------------------------------------------

d)
y --> 0 as t --> 0 is TRUE
The function is defined everywhere on the domain. --> TRUE

-------------------------------------------------------------

e)
Notice we have even degree, positive leading term
So, from end behavior, it GOES UP ON BOTH ENDS

So, y ---> +inf
as t ---> inf or -inf

Option A ----> ANS

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