Let f be a differentiable function with f(-3)= 1 and f'(−3)= 1 and f'(-3)= -5
Let the function g(x)=3[f(x)]^2
Write the equation of the line tangent to the graph of gg at the point where x= -3
First, how can you say that "f'(−3)= 1 and f'(-3)= -5" ?? Pick one. I pick 1
using the chain rule, we have g' = 6f f'
g'(-3) = 6f(-3) f'(-3) = 6(1)(-1) = -6
g(3) = 3(-3)^2 = -27
So now we have a point and a slope, giving us the equation
g+27 = -6(x+3)
To find the equation of the tangent line to the graph of g(x) at x = -3, we can follow these steps:
Step 1: Find the value of g(-3)
- Plug x = -3 into the function g(x) = 3[f(x)]^2.
- Since we know f(-3) = 1, substitute it into the equation: g(-3) = 3[f(-3)]^2 = 3(1)^2 = 3.
Step 2: Find the slope of the tangent line
- Calculate g'(-3), the derivative of g(x) at x = -3.
- Use the chain rule for differentiation. Recall that g(x) = 3[f(x)]^2, so the derivative of g(x) with respect to x is g'(x) = 6f(x)f'(x).
- Substitute x = -3 into the equation: g'(-3) = 6f(-3)f'(-3) = 6(1)(-5) = -30.
Step 3: Write the equation of the tangent line
- The equation of a line with slope m passing through point (x1, y1) is given by y - y1 = m(x - x1).
- Substitute the values x1 = -3, y1 = g(-3) = 3, and m = g'(-3) = -30 into the equation: y - 3 = -30(x - (-3)).
- Simplify the equation to get the final answer: y - 3 = -30(x + 3).
Therefore, the equation of the line tangent to the graph of g at x = -3 is y - 3 = -30(x + 3).
To find the equation of the tangent line to the graph of g(x) at x = -3, we need to find the slope of the tangent line and the point of tangency.
First, let's find the value of g(x) when x = -3:
g(-3) = 3[f(-3)]^2
Given that f(-3) = 1, we can substitute this value into the equation:
g(-3) = 3(1)^2
g(-3) = 3
So, the point of tangency on the graph of g(x) is (-3, 3).
Next, we need to find the derivative of g(x) to find the slope of the tangent line at x = -3.
Using the chain rule, the derivative of g(x) is:
g'(x) = 2[f(x)] * f'(x)
Given that f'(-3) = -5:
g'(x) = 2[f(x)] * -5
Now, let's find the value of g'(-3):
g'(-3) = 2[f(-3)] * f'(-3)
g'(-3) = 2(1) * (-5)
g'(-3) = -10
So, the slope of the tangent line to the graph of g(x) at x = -3 is -10.
Now we have the point of tangency (-3, 3) and the slope of the tangent line as -10. We can use the point-slope form of a linear equation to find the equation of the tangent line:
y - y1 = m(x - x1)
where m is the slope of the line and (x1, y1) is the point of tangency.
Substituting the values, we have:
y - 3 = -10(x - (-3))
Simplifying:
y - 3 = -10(x + 3)
y - 3 = -10x - 30
y = -10x - 30 + 3
y = -10x - 27
Therefore, the equation of the line tangent to the graph of g(x) at x = -3 is y = -10x - 27.