Question
find slope of tangent line to a curve of 3/sqrt(x)
Answers
Answer
use the formula of f(a+h)-f(a)/h
Answered by
GPT 3.5
To find the slope of the tangent line to the curve of \(f(x) = \frac{3}{\sqrt{x}}\), we need to find the derivative of the function \(f(x)\).
Using the power rule for differentiation, we have:
\[
\begin{aligned}
f'(x) &= \frac{d}{dx} \left(\frac{3}{\sqrt{x}}\right) \\
&= 3 \cdot \frac{d}{dx} \left(x^{-1/2}\right) \\
&= 3 \cdot \left(-\frac{1}{2}\right) \cdot x^{-1/2 - 1} \\
&= -\frac{3}{2}x^{-3/2}.
\end{aligned}
\]
Therefore, the slope of the tangent line to the curve \(f(x) = \frac{3}{\sqrt{x}}\) is \(-\frac{3}{2}x^{-3/2}\).
Using the power rule for differentiation, we have:
\[
\begin{aligned}
f'(x) &= \frac{d}{dx} \left(\frac{3}{\sqrt{x}}\right) \\
&= 3 \cdot \frac{d}{dx} \left(x^{-1/2}\right) \\
&= 3 \cdot \left(-\frac{1}{2}\right) \cdot x^{-1/2 - 1} \\
&= -\frac{3}{2}x^{-3/2}.
\end{aligned}
\]
Therefore, the slope of the tangent line to the curve \(f(x) = \frac{3}{\sqrt{x}}\) is \(-\frac{3}{2}x^{-3/2}\).
Answered by
GPT 3.5
Certainly! Using the formula for the slope of a secant line, we can approximate the slope of the tangent line to the curve of the function \(f(x) = \frac{3}{\sqrt{x}}\) using the formula \(\frac{f(a+h) - f(a)}{h}\), where \(a\) is a specific value and \(h\) is a small change in \(x\).
Let's proceed with finding this slope at a specific point. Let's choose the value \(a\) to be \(a = 4\) for example.
Step 1: Determine \(f(a)\) and \(f(a+h)\).
\(f(a) = f(4) = \frac{3}{\sqrt{4}} = \frac{3}{2}\)
\(f(a+h) = f(4+h) = \frac{3}{\sqrt{4+h}}\)
Step 2: Substitute the values into the formula to calculate the slope of the secant line:
Slope of the secant line = \(\frac{f(a+h) - f(a)}{h} = \frac{\frac{3}{\sqrt{4+h}} - \frac{3}{2}}{h}\)
Step 3: Simplify the expression:
\(\frac{\frac{3}{\sqrt{4+h}} - \frac{3}{2}}{h} = \frac{3}{2} \cdot \frac{2 - \sqrt{4+h}}{\sqrt{4+h} \cdot h}\)
Therefore, the slope of the secant line to the curve \(f(x) = \frac{3}{\sqrt{x}}\) at the point \(a = 4\) is \(\frac{3}{2} \cdot \frac{2 - \sqrt{4+h}}{\sqrt{4+h} \cdot h}\), where \(h\) represents a small change in \(x\). This value approximates the slope of the tangent line to the curve at the point \(x = 4\).
Let's proceed with finding this slope at a specific point. Let's choose the value \(a\) to be \(a = 4\) for example.
Step 1: Determine \(f(a)\) and \(f(a+h)\).
\(f(a) = f(4) = \frac{3}{\sqrt{4}} = \frac{3}{2}\)
\(f(a+h) = f(4+h) = \frac{3}{\sqrt{4+h}}\)
Step 2: Substitute the values into the formula to calculate the slope of the secant line:
Slope of the secant line = \(\frac{f(a+h) - f(a)}{h} = \frac{\frac{3}{\sqrt{4+h}} - \frac{3}{2}}{h}\)
Step 3: Simplify the expression:
\(\frac{\frac{3}{\sqrt{4+h}} - \frac{3}{2}}{h} = \frac{3}{2} \cdot \frac{2 - \sqrt{4+h}}{\sqrt{4+h} \cdot h}\)
Therefore, the slope of the secant line to the curve \(f(x) = \frac{3}{\sqrt{x}}\) at the point \(a = 4\) is \(\frac{3}{2} \cdot \frac{2 - \sqrt{4+h}}{\sqrt{4+h} \cdot h}\), where \(h\) represents a small change in \(x\). This value approximates the slope of the tangent line to the curve at the point \(x = 4\).