Question
given d(x) =16x^2 -5 find d ^-1(x)
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Answer
d(x) =16x^2 -5 find d ^-1(x)
Answered by
GPT-4 Turbo
To find the inverse function \( d^{-1}(x) \), we need to swap the input \( x \) and the output \( y \) in the original function and then solve for the new input. Let's denote the original function \( d(x) \) as \( y \).
Given the function:
\[ d(x) = y = 16x^2 - 5 \]
Swap the variables:
\[ x = 16y^2 - 5 \]
Next, we solve for \( y \) in terms of \( x \):
\[ x + 5 = 16y^2 \]
\[ \frac{x + 5}{16} = y^2 \]
We need to take the square root of both sides to solve for \( y \). Because we're looking for a function, we should be aware that there might be two solutions (a positive and a negative square root). However, in most function cases, we select the principal (positive) square root to ensure the inverse is also a function. If we are considering the real-valued function, and keeping in mind that the original function \( d(x) \) suggests \( x \) was non-negative (since it was in a squared term), we will return the principal root here:
\[ y = \pm\sqrt{\frac{x + 5}{16}} \]
\[ y = \pm\frac{\sqrt{x + 5}}{4} \]
As the original function is a parabola opening upwards (because the coefficient of \( x^2 \) is positive) and the vertex is at a minimum point, the original function is increasing for all \( x \), meaning that for each \( x \) there is one unique \( y \). Therefore, the proper inverse function will also be a function (passing the horizontal line test), meaning we can choose the positive root for the inverse:
\[ y = \frac{\sqrt{x + 5}}{4} \]
Now we can write the inverse function \( d^{-1}(x) \):
\[ d^{-1}(x) = \frac{\sqrt{x + 5}}{4} \]
Keep in mind that the domain of this inverse function will be restricted to \( x \geq -5 \) because the original function's range only includes outputs where \( y \geq -5 \). This restriction ensures that \( x + 5 \) is non-negative, which is necessary for the square root to be defined in the real number system.
Given the function:
\[ d(x) = y = 16x^2 - 5 \]
Swap the variables:
\[ x = 16y^2 - 5 \]
Next, we solve for \( y \) in terms of \( x \):
\[ x + 5 = 16y^2 \]
\[ \frac{x + 5}{16} = y^2 \]
We need to take the square root of both sides to solve for \( y \). Because we're looking for a function, we should be aware that there might be two solutions (a positive and a negative square root). However, in most function cases, we select the principal (positive) square root to ensure the inverse is also a function. If we are considering the real-valued function, and keeping in mind that the original function \( d(x) \) suggests \( x \) was non-negative (since it was in a squared term), we will return the principal root here:
\[ y = \pm\sqrt{\frac{x + 5}{16}} \]
\[ y = \pm\frac{\sqrt{x + 5}}{4} \]
As the original function is a parabola opening upwards (because the coefficient of \( x^2 \) is positive) and the vertex is at a minimum point, the original function is increasing for all \( x \), meaning that for each \( x \) there is one unique \( y \). Therefore, the proper inverse function will also be a function (passing the horizontal line test), meaning we can choose the positive root for the inverse:
\[ y = \frac{\sqrt{x + 5}}{4} \]
Now we can write the inverse function \( d^{-1}(x) \):
\[ d^{-1}(x) = \frac{\sqrt{x + 5}}{4} \]
Keep in mind that the domain of this inverse function will be restricted to \( x \geq -5 \) because the original function's range only includes outputs where \( y \geq -5 \). This restriction ensures that \( x + 5 \) is non-negative, which is necessary for the square root to be defined in the real number system.
Answered by
GPT-4 Turbo
To find the inverse function \( d^{-1}(x) \) of the function \( d(x) = 16x^2 - 5 \), we have to swap the output \( d(x) \) with the input \( x \) and solve for the new input.
Let's start with the function:
\[ d(x) = 16x^2 - 5 \]
Now, swap \( d(x) \) with \( x \):
\[ x = 16y^2 - 5 \]
Next, solve for \( y \) in terms of \( x \):
\[ x + 5 = 16y^2 \]
\[ \frac{x + 5}{16} = y^2 \]
To find \( y \), we take the square root of both sides. However, since we want the inverse to also be a function, we need to consider the domain and range of the original function. The original function \( d(x) \) is a parabola that opens upwards and is shifted 5 units down. This means the parabola has a minimum point at \( y = -5 \) and \( d(x) \) is increasing for \( x > 0 \) and decreasing for \( x < 0 \).
Since the inverse function should be a reflection across the line \( y = x \), the domain of the original function becomes the range of the inverse, and the range of the original function becomes the domain of the inverse. As \( d(x) \) is only ever increasing, we can take only the positive square root for the inverse to ensure it is also a function.
\[ y = \sqrt{\frac{x + 5}{16}} \]
\[ y = \frac{\sqrt{x + 5}}{4} \]
Thus the inverse function \( d^{-1}(x) \) is:
\[ d^{-1}(x) = \frac{\sqrt{x + 5}}{4} \]
where \( x \geq -5 \) to ensure that the square root is defined over the real numbers. This condition corresponds to the fact that the range of \( d(x) \) begins at \(-5\) and goes to infinity.
Let's start with the function:
\[ d(x) = 16x^2 - 5 \]
Now, swap \( d(x) \) with \( x \):
\[ x = 16y^2 - 5 \]
Next, solve for \( y \) in terms of \( x \):
\[ x + 5 = 16y^2 \]
\[ \frac{x + 5}{16} = y^2 \]
To find \( y \), we take the square root of both sides. However, since we want the inverse to also be a function, we need to consider the domain and range of the original function. The original function \( d(x) \) is a parabola that opens upwards and is shifted 5 units down. This means the parabola has a minimum point at \( y = -5 \) and \( d(x) \) is increasing for \( x > 0 \) and decreasing for \( x < 0 \).
Since the inverse function should be a reflection across the line \( y = x \), the domain of the original function becomes the range of the inverse, and the range of the original function becomes the domain of the inverse. As \( d(x) \) is only ever increasing, we can take only the positive square root for the inverse to ensure it is also a function.
\[ y = \sqrt{\frac{x + 5}{16}} \]
\[ y = \frac{\sqrt{x + 5}}{4} \]
Thus the inverse function \( d^{-1}(x) \) is:
\[ d^{-1}(x) = \frac{\sqrt{x + 5}}{4} \]
where \( x \geq -5 \) to ensure that the square root is defined over the real numbers. This condition corresponds to the fact that the range of \( d(x) \) begins at \(-5\) and goes to infinity.
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