let's recall the remainder theorem.
we know that (x-1) is a factor, that means x -1 = 0 or x = 1.
since we know that (x-1) is a factor, then dividing the polynomial by it will give us a remainder of 0, which correlates with saying that f(1) = 0, in this case, so we can simply plug in "1" as the argument, knowing it gives 0.
![f(x)=3x^3+kx-11\\\\[-0.35em]~\dotfill\\\\\stackrel{0}{f(1)}=3(1)^3+k(1)-11\implies \stackrel{f(1)}{0}=3+k-11\implies 0=-8+k\implies 8=k](https://tex.z-dn.net/?f=f%28x%29%3D3x%5E3%2Bkx-11%5C%5C%5C%5C%5B-0.35em%5D~%5Cdotfill%5C%5C%5C%5C%5Cstackrel%7B0%7D%7Bf%281%29%7D%3D3%281%29%5E3%2Bk%281%29-11%5Cimplies%20%5Cstackrel%7Bf%281%29%7D%7B0%7D%3D3%2Bk-11%5Cimplies%200%3D-8%2Bk%5Cimplies%208%3Dk)
Answer:
Step-by-step explanation:
Isolate the variable by dividing each side by factors that don't contain the variable.
Exact Form: x=-45
x
=
−
5
4
Decimal Form:x=-1.25
x
=
−
1.25
Mixed Number Form:x=-14
x
=
−
1
1
4
Answer:
y=9
Step-by-step explanation:
Slope: 9-9/-2-15=0
Therefore, y=0x+b
Plug values in: 9=0(-2)+b
0+b=9
b=9
Therefore, equation is y=9
Y = e^tanx - 2
To find at which point it crosses x axis we state that y= 0
e^tanx - 2 = 0
e^tanx = 2
tanx = ln 2
tanx = 0.69314
x = 0.6061
to find slope at that point first we need to find first derivative of funtion y.
y' = (e^tanx)*1/cos^2(x)
now we express x = 0.6061 in y' and we get:
y' = k = 2,9599
