Finding k for three real solutions
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The function $f(x) = x^3 - x^2 - x - \frac{11}{4}$ is graphed in the xy plane above. If k is a constant such that the equation $f(x) = k$ has three real solutions, which of the following could be the value of k? A) 2 B) 0 C) -2 D) -3
This question includes visual content: A Cartesian coordinate system showing the graph of a cubic function. The y-axis ranges from -5 to 5, and the x-axis ranges from -5 to 5. The curve enters from the third quadrant, has a local maximum near x = -0.5, y = -2.5, a local minimum near x = 1, y = -3.5, and then increases upwards, crossing the x-axis between 2 and 3.
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Hi Emir, let's solve this problem together by finding the value of k for which the equation has three real solutions.
Graphical Analysis of $f(x) = k$
The equation f of x equals k represents the intersection of the curve y equals f of x with a horizontal line y equals k.
For $f(x) = k$ to have exactly three real solutions, the horizontal line $y = k$ must intersect the graph at exactly three points.
Looking at the graph, this intersection only happens when the horizontal line lies strictly between the local minimum and the local maximum of the function.
Let's find the exact values of the local minimum and maximum using calculus. We start by writing our function.
Finding Critical Points
We take the first derivative with respect to x and set it equal to zero.
We can factor this quadratic equation to solve for the critical points.
This gives us two critical points: x equals negative one third and x equals one.
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