Now we plug in and solve.
One should also note that, because H = 3R, = 3 . Thus, after we found in part 2, we
merely had to multiply it by 3 to find the answer for part 3.
- Let f be the function given by y = f(x) = 2x^4 − 4x^2 + 1.
(a) Find an equation of the line tangent to the graph at (−2, 17).
In order to find the equation of a tangent line at a particular point, we need to take the
derivative of the function and plug in the x- and y-values at that point to give us the slope of the
line.
Step 1: The derivative is f′(x) = 8x^3 − 8x. If we plug in x = −2, we get
f′(−2) = 8(−2)^3 − 8(−2) = −48
This is the slope m.
Step 2: Now we use the slope-intercept form of the equation of a line, y − y 1 = m(x − x 1 ), and
plug in the appropriate values of x, y, and m.
y − 17 = −48(x + 2)
If we simplify this we get y = −48x − 79.
(b) Find the x- and y-coordinates of the relative maxima and relative minima. Verify your
answer.
If we want to find the maxima/minima, we need to take the derivative and set it equal to zero.
The values that we get are called critical points. We will then test each point to see if it is a
maximum or a minimum.
