Given the function \(y(x)=(x+3)(x-2)\), for \(-4<x<4\). What is the value of \(x\) at which the function has a minimum?
Show Hint
Solution and Explanation
Step 1: Expand the quadratic.
\[
y(x) = (x+3)(x-2) = x^2 + 3x - 2x - 6 = x^2 + x - 6.
\]
Step 2: Nature of parabola.
Coefficient of \(x^2\) is positive (\(+1\)), so it is an upward-opening parabola.
Hence, it has a single global minimum at its vertex.
Step 3: Find vertex.
For quadratic \(y=ax^2+bx+c\), vertex \(x=-\tfrac{b}{2a}\).
Here \(a=1, b=1\):
\[
x=-\frac{1}{2(1)}=-\frac{1}{2}.
\]
Step 4: Check domain.
Domain is \(-4<x<4\). Since \(-0.5\) lies inside, that is the minimum point.
Final Answer:
\[
\boxed{-\tfrac{1}{2}}
\]
Top Questions on Function
- Given that the Laplace transforms of \( J_0(x), J_0'(x), \) and \( J_0''(x) \) exist, where \( J_0(x) \) is the Bessel function. Let \( Y = Y(s) \) be the Laplace transform of the Bessel function \( J_0(x) \). Then, which one of the following is TRUE?
Let \( X = \{ f \in C[0,1] : f(0) = 0 = f(1) \} \) with the norm \( \|f\|_\infty = \sup_{0 \leq t \leq 1} |f(t)| \), where \( C[0,1] \) is the space of all real-valued continuous functions on \( [0,1] \).
Let \( Y = C[0,1] \) with the norm \( \|f\|_2 = \left( \int_0^1 |f(t)|^2 \, dt \right)^{\frac{1}{2}} \). Let \( U_X \) and \( U_Y \) be the closed unit balls in \( X \) and \( Y \) centered at the origin, respectively. Consider \( T: X \to \mathbb{R} \) and \( S: Y \to \mathbb{R} \) given by
\[ T(f) = \int_0^1 f(t) \, dt \quad \text{and} \quad S(f) = \int_0^1 f(t) \, dt. \]
Consider the following statements:
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S2: \( \sup |S(f)| \) is attained at a point of \( U_Y \).
Then, which one of the following is correct?Let \( g(x, y) = f(x, y)e^{2x + 3y} \) be defined in \( \mathbb{R}^2 \), where \( f(x, y) \) is a continuously differentiable non-zero homogeneous function of degree 4. Then,
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\[ f(x, y) = \frac{x^2 - y^2}{x^2 + y^2} + x \sin \left( \frac{1}{x^2 + y^2} \right). \]
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S3: \( \lim_{(x, y) \to (0, 0)} f(x, y) \) exists.
Then, which of the following is/are correct?
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