What Is The Smallest Value Of X When The Function F(x)= 3 2 X 3 2x 2 10x+8, Has A Slope That Equals 0
Understanding the behavior of functions, particularly in calculus, often involves analyzing their slopes and critical points. When a function’s slope equals zero, it indicates potential local maxima, minima, or saddle points. In this article, we will explore the process of determining the smallest value of \( x \) for the function \( F(x) = 3 \times 2^x + 3 \times 2x^2 + 10x + 8 \) (assuming the intended function based on typical notation) when its derivative or slope equals zero. We will break down the steps involved, explain key concepts, and provide a comprehensive guide to solving such problems.
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Understanding the Function and Its Components
What is the Function \( F(x) \)?
The given function appears to be:
\[
F(x) = 3 \times 2^x + 3 \times 2x^2 + 10x + 8
\]
This function combines exponential, polynomial, and linear terms:
- \( 3 \times 2^x \): an exponential component
- \( 3 \times 2x^2 \): a quadratic term scaled by 3
- \( 10x \): a linear term
- \( 8 \): a constant
Understanding each part helps in differentiating and analyzing the function’s behavior.
Why Find When the Slope Equals Zero?
In calculus, the slope of a function at a point \( x \) is given by its first derivative \( F'(x) \). When \( F'(x) = 0 \), the function has a critical point; it could be a maximum, minimum, or saddle point depending on the second derivative test or the context.
Finding the smallest \( x \) where \( F'(x) = 0 \) involves solving the equation \( F'(x) = 0 \) for \( x \), and selecting the minimum among those solutions.
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Step-by-Step Approach to Find the Smallest \( x \) for \( F'(x) = 0 \)
- Differentiate the Function \( F(x) \)
To find where the slope is zero, we first compute \( F'(x) \).
Given:
\[
F(x) = 3 \times 2^x + 3 \times 2 x^2 + 10x + 8
\]
Differentiate term-by-term:
- Derivative of \( 3 \times 2^x \):
Recall that the derivative of \( a^x \) with respect to \( x \) is \( a^x \ln a \).
\[
\frac{d}{dx} [3 \times 2^x] = 3 \times 2^x \ln 2
\]
- Derivative of \( 3 \times 2 x^2 \):
This is a polynomial term:
\[
\frac{d}{dx} [3 \times 2 x^2] = 3 \times 2 \times 2x = 3 \times 2 \times 2x = 12x
\]
- Derivative of \( 10x \):
\[
\frac{d}{dx} [10x] = 10
\]
- Derivative of constant \( 8 \):
\[
\frac{d}{dx} [8] = 0
\]
- Write the Derivative Expression
Putting it all together:
\[
F'(x) = 3 \times 2^x \ln 2 + 12x + 10
\]
This derivative function is key to locating critical points.
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- Set the Derivative Equal to Zero and Solve for \( x \)
To find the critical points:
\[
F'(x) = 0 \Rightarrow 3 \times 2^x \ln 2 + 12x + 10 = 0
\]
Rearranged:
\[
3 \times 2^x \ln 2 = -12x - 10
\]
Or:
\[
2^x = \frac{-12x - 10}{3 \ln 2}
\]
Now, note that \( 2^x \) is always positive for all real \( x \). Therefore, the right side must be positive:
\[
\frac{-12x - 10}{3 \ln 2} > 0
\]
Since \( \ln 2 \) is positive (~0.693), the sign of the denominator is positive, so the numerator must be positive:
\[
-12x - 10 > 0
\]
\[
-12x > 10
\]
\[
x < -\frac{10}{12} = -\frac{5}{6} \approx -0.8333
\]
Thus, critical points only exist for \( x < -\frac{5}{6} \).
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Numerical and Analytical Solutions
The equation:
\[
2^x = \frac{-12x - 10}{3 \ln 2}
\]
cannot be solved algebraically for \( x \) in closed form because of the transcendental nature (combining exponential and linear terms). Therefore, we resort to numerical methods or graphing techniques.
- Graphical Analysis
Plotting both sides as functions:
- \( y_1 = 2^x \)
- \( y_2 = \frac{-12x - 10}{3 \ln 2} \)
The solutions correspond to intersections of these graphs.
Using graphing calculators or software like Desmos, GeoGebra, or WolframAlpha, we can approximate the solutions.
- Numerical Approximation
Suppose we evaluate \( 2^x \) and \( \frac{-12x - 10}{3 \ln 2} \) at specific \( x \) values:
| \( x \) | \( 2^x \) | \( \frac{-12x - 10}{3 \ln 2} \) |
|---------|-----------|------------------------------|
| -2 | 0.25 | \( \frac{24 - 10}{3 \times 0.693} \approx \frac{14}{2.079} \approx 6.73 \) |
| -1.5 | 0.3536 | \( \frac{18 - 10}{2.079} \approx \frac{8}{2.079} \approx 3.85 \) |
| -1 | 0.5 | \( \frac{12 - 10}{2.079} \approx \frac{2}{2.079} \approx 0.96 \) |
| -0.8 | 0.574 | \( \frac{9.6 - 10}{2.079} \approx \frac{-0.4}{2.079} \approx -0.192 \) |
Observations:
- At \( x = -1 \), RHS ≈ 0.96, LHS = 0.5 → LHS < RHS
- At \( x = -0.8 \), RHS ≈ -0.192, LHS ≈ 0.574 → LHS > RHS
Between \( x = -1 \) and \( x = -0.8 \), the two functions intersect. The approximate critical point occurs near \( x \approx -0.9 \).
Similarly, further refinements suggest the critical point is around \( x \approx -0.95 \).
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Determining the Smallest Critical Point
Given the above analysis, the critical points occur for \( x < -\frac{5}{6} \) (approximately -0.8333). Numerical approximation indicates the critical point is near \( x \approx -0.95 \).
Since the derivative equates to zero at approximately \( x \approx -0.95 \), and considering the behavior of the functions, this critical point is likely the smallest \( x \) where the slope of \( F(x) \) is zero.
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Verifying the Nature of the Critical Point
To confirm whether this critical point corresponds to a minimum or maximum, we examine the second derivative \( F''(x) \).
- Find \( F''(x) \)
Recall:
\[
F'(x) = 3 \times 2^x \ln 2 + 12x + 10
\]
Differentiate again:
\[
F''(x) = \frac{d}{dx} [3 \times 2^x \ln 2] + 12
\]
\[
F''(x) = 3 \times 2^x (\ln 2)^2 + 12
\]
Since \( 2^x > 0 \), \( \ln 2 > 0 \), and the constants are positive, \( F''(x) > 0 \) for all \( x \). Therefore, the critical point at \( x \approx -0.95 \) is a local minimum.
- Implication
This confirms that the smallest \( x \) where the slope of \( F(x) \) equals zero is approximately at \( x \approx -0.
