To find the equation of the tangent line to the graph of [tex]\( f(x) = -\frac{1}{3} x^3 \)[/tex] at [tex]\( x = -1 \)[/tex], follow these steps:
1. Calculate the derivative of [tex]\( f(x) \)[/tex] to find the slope of the tangent line:
The function is [tex]\( f(x) = -\frac{1}{3} x^3 \)[/tex].
The derivative [tex]\( f'(x) \)[/tex] is found using the power rule [tex]\( \frac{d}{dx} x^n = n x^{n-1} \)[/tex].
[tex]\[
f'(x) = -\frac{1}{3} \cdot 3 x^2 = -x^2
\][/tex]
2. Evaluate the derivative at [tex]\( x = -1 \)[/tex] to get the slope of the tangent line:
[tex]\[
f'(-1) = -(-1)^2 = -1
\][/tex]
So, the slope of the tangent line at [tex]\( x = -1 \)[/tex] is [tex]\( -1 \)[/tex].
3. Find the function value at [tex]\( x = -1 \)[/tex] to get the point of tangency:
[tex]\[
f(-1) = -\frac{1}{3} (-1)^3 = -\frac{1}{3} \cdot (-1) = \frac{1}{3}
\][/tex]
Thus, the point of tangency is [tex]\( (-1, \frac{1}{3}) \)[/tex].
4. Use the point-slope form of the equation of a line to find the equation of the tangent line:
The point-slope form is [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is the point of tangency.
Plugging in the values [tex]\( m = -1 \)[/tex], [tex]\( x_1 = -1 \)[/tex], and [tex]\( y_1 = \frac{1}{3} \)[/tex]:
[tex]\[
y - \frac{1}{3} = -1 (x - (-1))
\][/tex]
Simplifying, we get:
[tex]\[
y - \frac{1}{3} = -1 (x + 1)
\][/tex]
[tex]\[
y - \frac{1}{3} = -x - 1
\][/tex]
[tex]\[
y = -x - 1 + \frac{1}{3}
\][/tex]
[tex]\[
y = -x - \frac{3}{3} + \frac{1}{3}
\][/tex]
[tex]\[
y = -x - \frac{2}{3}
\][/tex]
So, the equation of the tangent line in slope-intercept form is:
[tex]\[
y = -x - \frac{2}{3}
\][/tex]
