Answer:
286
Step-by-step explanation:
Answer:
(-∞, -1-(2/3)√2] ∪ [-1+(2/3)√2, ∞)
Step-by-step explanation:
To make it easier to differentiate, we'll rewrite the function as ...
h(x) = 2/(x-5) -1/(x-2)
Then the derivative is ...
h'(x) = -2/(x-5)^2 +1/(x-2)^2
This will be zero when ...
-2(x-2)^2 +(x-5)^2 = 0
-2(x^2 -4x +4) +(x^2 -10x +25) = 0
-x^2 -2x +17 = 0
x^2 +2x +1 = 17 +1
x +1 = ±√18 = ±3√2
x = -1 ±3√2
The values of the function at these locations are ...
h(-1-3√2) = -1 +(2/3)√2 ≈ -1.9428
h(-1+3√2) = -1 -(2/3)√2 ≈ -0.0572
Then the range of h(x) is ...
(-∞, -1-(2/3)√2] ∪ [-1+(2/3)√2, ∞)
Try going with this method for that question ⬇⬇⬇⬇
k = 273 + C
Answer:
vertical asymptote at x = 7
horizontal asymptote at y = 6
Step-by-step explanation:
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.
solve:
x
−
7
=
0
⇒
x
=
7 is the asymptote
Horizontal asymptotes occur as
lim
,f(x)→
c(a constant)
x
→
±
∞
divide terms on numerator/denominator by x
f
(
x
)
=
5/
x
+6=
5
/x
+
6
x/
x
−
7
/x 1
−
7/
x
as
x
±
∞
,
f
(
x
)
→
0
+6
1
−
0
⇒
y
=
6 is the asymptote
graph{((5)/(x-7))+6 [-20, 20, -10, 10]}
Hey there!
You could
"
" into your terms
Next, you have to
your 



- The "
" stays the same since it doesn't have a like term in this equation 
Good luck on your assignment and enjoy your day!
~