Suppose that f(x) and k(x) are differentiable everywhere and that f(3) = 5, f'(3) = 7, k(3) = 6, and k'(3) = 9. Find an equation of a tangent line to 3(f(x))^2 at x = 3.
So the derivative of 3(f(x))^2 is 6f(x)*f'(x), and 6f(3)*f'(3) is 210. But for the tangent line, should I use f(3) = 5 for the y-value, or should I use 3(f(3))^2 = 75 instead?
the slope of 3f^2 is 6f f' = 6*5*7=210
So far, so good.
3f^2(3) = 3*5^2 = 75
So, you want the line through (3,75) with slope=210
y-75 = 210(x-3)
your surmise is correct
To find the equation of a tangent line to 3(f(x))^2 at x = 3, we need both the value of f(x) and its derivative f'(x) at x = 3.
Given that f(3) = 5 and f'(3) = 7, we can use these values to find the slope of the tangent line. The slope of the tangent line is equal to the derivative of the function at the point of tangency.
Now, let's calculate the slope:
m = f'(3) = 7
Using the point-slope form of a linear equation, the equation of the tangent line can be given by:
y - y1 = m(x - x1)
Plugging in the known values, we have:
y - 5 = 7(x - 3)
Next, we simplify and rewrite the equation in slope-intercept form (y = mx + b):
y - 5 = 7x - 21
Combining like terms, we get:
y = 7x - 16
Therefore, the equation of the tangent line to 3(f(x))^2 at x = 3 is:
y = 7x - 16.
To address your second question, since we are finding the equation for the tangent line to 3(f(x))^2, we should use f(3) = 5 for the y-coordinate. The expression 3(f(3))^2 = 75 would be relevant if we were finding the value of the function itself, not the equation of the tangent line.