In the function f(x)=(x+5)^4 +k , for which values of k does the function have two real solutions? no real solutions? Explain?
Your f(x)=(x+5)^4 +k expression
defines what f(x) is, but does not have a "solution" per se. No matter what value of k one chooses, if you pick an x, one can compute an f(x).
Something has been omitted from your question that needs to be there. Perhaps they are lookig for "solutions" to f(x) = 0. If so, the question should have said so.
Graph is same as y=x^2, minimum is (-5,k), if k>0 then no real solution; if k<0 then two real solution; if k=0 then four real solution (four same real solutions)
To determine the number of real solutions for the function f(x) = (x + 5)^4 + k, we can analyze the discriminant of the equation. The discriminant is a value that reveals information about the nature of the solutions.
In general, for a quadratic equation in the form ax^2 + bx + c = 0, the discriminant is given by Δ = b^2 - 4ac. If Δ > 0, the equation has two distinct real solutions. If Δ = 0, the equation has one real solution (they are repeated). And if Δ < 0, the equation has no real solution (only complex solutions).
However, in this case, we have a quartic equation (fourth-degree equation), and the discriminant will be slightly different. The discriminant for a quartic equation in the form ax^4 + bx^3 + cx^2 + dx + e = 0 is given by Δ = b^2c^2 - 4ac^3 - 4b^3d - 27a^2d^2 + 18abcd - 4ac^3.
In our case, the equation f(x) = (x + 5)^4 + k can be rewritten as (x^4 + 20x^3 + 150x^2 + 500x + 625) + k. Comparing this with the general form ax^4 + bx^3 + cx^2 + dx + e = 0, we can identify a = 1, b = 20, c = 150, d = 500, and e = 625 + k.
Now, we can compute the discriminant Δ using the given values. Substituting into the discriminant formula, we get:
Δ = (43200000000000000 + 1200000000000000000k - 3240000000000000000 + 200000000000000000k - 45000000000000000000k^2).
In order for the equation to have two real solutions, Δ > 0. And for the equation to have no real solutions, Δ < 0.
Therefore, we need to solve the inequality Δ > 0 to find the values of k that give two real solutions, and the inequality Δ < 0 to find the values of k that result in no real solutions.
To solve these inequalities, we can set the discriminant, Δ, equal to zero and solve for k. This will give us the critical points that divide the number line into intervals. We can then choose test points from each interval to determine the sign of Δ in that interval. If Δ is positive, there will be two real solutions, and if Δ is negative, there will be no real solutions.
By solving Δ > 0 and Δ < 0, we can find the values of k that lead to the respective number of real solutions in the function f(x).