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3 years ago

What is the value of k in the function ƒ(x) = 112 - kx if ƒ(-3) = 121? k =

Mathematics

2 answers:

masha68 [24]3 years ago

5 0

Plug -3 into the function to get 112+3k=f(-3)=121.
So now can calculate k: 112+3k=121, 3k=9, k=3.

Oduvanchick [21]3 years ago

5 0

Ok to solve for k, take the function f(x) = 112 - kx plug in the x value, so that would be f(-3) now we know that f(-3) = 121 so how to solve for k is quite easy. In replace f(-3) with 121 and x with -3
so 121 = 112 - k(-3)
solve for k
subtract 112 from both sides
so that would give you 9 = -k(-3)
use the distributive property on -k(-3)
so that would be 3k = 9
divide both sides by 3
which gives you the answer k=3

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Which of the values shown are potential roots of f(x) = 3x3 – 13x2 – 3x + 45? Select all that apply.

galina1969 [7]

Answer:

All potential roots are 3,3 and $-\frac{5}{3}$ .

Step-by-step explanation:

Potential roots of the polynomial is all possible roots of f(x).

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Using rational root theorem test. We will find all the possible or potential roots of the polynomial.

p=All the positive/negative factors of 45

q=All the positive/negative factors of 3

$p=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45$

$q=\pm 1,\pm 3$

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$\frac{p}{q}=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45,\pm \frac{1}{3},\pm \frac{5}{3}$

Now we check each rational root and see which are possible roots for given function.

$f(1)= 3\times 1^3-13\times 1^2-3\times 1+45\Rightarrow 32\neq 0$

$f(-1)= 3\times (-1)^3-13\times (-1)^2-3\times (-1)+45\Rightarrow \neq 32$

$f(-3)= 3\times (-3)^3-13\times (-3)^2-3\times (-3)+45\Rightarrow \neq -144$

$f(3)= 3\times (3)^3-13\times (3)^2-3\times (3)+45\Rightarrow =0\\\\ \therefore x=3\text{ Potential roots of function}$

Similarly, we will check for all value of p/q and we get

$f(-5/3)=0$

Thus, All potential roots are 3,3 and $-\frac{5}{3}$ .

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