Find the equation of the line tangent to the function f(x)= square root of (x^3 + 4x) through the point (2,4)
f(x) = √(x^3+4x)
f'(x) = (3x^2+4) / 2√(x^3+4x)
f'(2) = 16/8 = 2
So, now you have a point and a slope. The line is
y-4 = 2(x-2)
y = 2x
See the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D%E2%88%9A%28x^3%2B4x%29%2C+y%3D2x+for+x%3D-1..3
To find the equation of the line tangent to the function \( f(x) = \sqrt{x^3 + 4x} \) through the point (2,4), we need to find the slope of the tangent line first. The slope of the tangent line can be found by taking the derivative of the function and evaluating it at the given point.
Step 1: Find the derivative of the function \( f(x) = \sqrt{x^3 + 4x} \).
To do this, we can use the chain rule. Let's first rewrite the function as \( f(x) = (x^3 + 4x)^{1/2} \):
\( f'(x) = \frac{1}{2}(x^3 + 4x)^{-1/2} \cdot (3x^2 + 4) \)
Step 2: Evaluate the derivative at x = 2 to find the slope of the tangent line.
Substituting x = 2 into the derivative formula:
\( f'(2) = \frac{1}{2}(2^3 + 4 \cdot 2)^{-1/2} \cdot (3 \cdot 2^2 + 4) \)
Simplifying further:
\( f'(2) = \frac{1}{2}(8 + 8)^{-1/2} \cdot (12 + 4) \)
\( f'(2) = \frac{1}{2}(16)^{-1/2} \cdot 16 \)
\( f'(2) = \frac{1}{2 \sqrt{16}} \cdot 16 \)
\( f'(2) = \frac{1}{2 \cdot 4} \cdot 16 \)
\( f'(2) = \frac{1}{8} \cdot 16 \)
\( f'(2) = 2 \)
The slope of the tangent line is 2.
Step 3: Write the equation of the tangent line using the point-slope form.
The point-slope form of a line is given by:
\( y - y_1 = m(x - x_1) \)
where m is the slope and (x1, y1) is a point on the line.
Substituting m = 2 and (x1, y1) = (2, 4) into the equation:
\( y - 4 = 2(x - 2) \)
Simplifying:
\( y - 4 = 2x - 4 \)
\( y = 2x \)
Therefore, the equation of the line tangent to the function \( f(x) = \sqrt{x^3 + 4x} \) through the point (2, 4) is \( y = 2x \).