What is the largest value of $k$ such that the equation $6x - x^2 = k$ has at least one real solution?

Mathematics

2 answers:

podryga [215]2 years ago

5 0

Step-by-step explanation:

6x - x² = k

=> x² - 6x + k = 0,

For there to be at least 1 real solution,

The discriminant b² - 4ac must be at least 0.

We have (-6)² - 4(1)(k) >= 0.

Therefore 36 - 4k >= 0, k <= 9.

Hence the largest value of k is 9.

amid [387]2 years ago

3 0

Answer:

Step-by-step explanation:

Rearranging, this is the equation x^2 - 6x = -k. Completing the square, we add 9 to both sides, and x^2 - 6x + 9 = 9 - k. Factoring the left side, we have (x - 3)^2 = 9 - k. In order for this equation to have at least one real solution, the right-hand side must be nonnegative. Thus, we have 9 - k $\geq$ 0 and k $\leq$ 9. Thus, the largest possible value of k is 9.

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Snezhnost [94]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

If F(x) = 2 - xand w(x) = x - 2, what is the range of (w•p(x)?

hoa [83]

Answer:

Th Range is [0, -∞)

Step-by-step explanation:

f(x) = 2 - x

w(x) = x - 2

We want to find the range of (f * w)(x).

First, we need to find (f * w)(x), which is the multiplication of the function f(x) and the function w(x). Lets use algebra to find (f * w)(x):

$(f*w)(x)=(2-x)(x-2)\\=2x-4-x^2+2x\\=-x^2+4x-4$

This is a quadratic function (U shaped), or a parabola. The graph is attached.

The range is the set of y-values for which the function is defined.

We see from the graph that the parabola is upside down and the highest value is y = 0 and lowest goes towards negative infinity. So the range is from 0 to negative infinity. Or,

0 < y < ∞

In interval notation, that would be:

[0, -∞)