Let f be a differentiable function such that f(3) = 2 and f'(3) = 5. If the tangent line to the graph of f at x = 3 is used to find an approximation to a zero of f, that approximation is?
So confused
y=mx+b f'(3)=5 means m=5
y=5x+b but f(3)=2 means
2=5*3+b, or b= -13
tangent line y= 5x-13
so the zero is 0=5x-13 x=13/5
That approximation is the intersection of
the tangent and X-axis.
The equation of the tangent:
y-2=5(x-3)
If y=0 then x=13/5
nice work, Mgraph.
Oh, don't be confused! I'm here to help with a smile. π Let's solve this problem together.
The equation of a tangent line can be written in the form of y = mx + b, where m represents the slope of the line. In this case, since the tangent line is to the graph of f at x = 3, we know that the slope of the tangent line is given by f'(3), which is 5 in this case.
To find the equation of the tangent line, we also need the y-coordinate of the point of tangency. We're given that f(3) = 2, so the point of tangency is (3, 2).
Now, we have a point on the line (3, 2) and the slope of the line (5). We can use the point-slope form of a line to find the equation of the tangent line:
y - 2 = 5(x - 3)
Expanding the equation:
y - 2 = 5x - 15
Simplifying further:
y = 5x - 13
Now, we want to find an approximation to a zero of f, which means we're looking for an x-value when y is equal to zero. So, we simply set y = 0 in the equation:
0 = 5x - 13
Solving for x:
5x = 13
x = 13/5
Therefore, the approximation to a zero of f, based on the tangent line, is x = 13/5.
I hope that clears things up! If you have any more questions, feel free to ask, and I'll be here with a joke and an answer. π
To approximate the zero of the function using the tangent line, we can make use of the concept of linear approximation. The tangent line to the graph of a differentiable function at a given point can be used to approximate the behavior of the function near that point.
In this case, the tangent line to the graph of f at x = 3 is given by the equation:
y = f'(3)(x - 3) + f(3)
where f'(3) represents the derivative of f at x = 3.
We are given that f(3) = 2 and f'(3) = 5. Plugging these values into the equation of the tangent line, we get:
y = 5(x - 3) + 2
Now, to find an approximation for the zero of f, we need to find the x-value for which y equals zero. We set y = 0 and solve for x:
0 = 5(x - 3) + 2
Simplifying the equation:
0 = 5x - 15 + 2
0 = 5x - 13
Now, we isolate x:
5x = 13
x = 13/5
So, the approximation for the zero of f, using the tangent line, is x = 13/5 or 2.6.
Remember that this approximation is only valid near x = 3, as it relies on the behavior of the function around that point. For more accurate approximations, numerical methods like Newton's method might be more appropriate.
To find an approximation to a zero of the function f using the tangent line at x = 3, we can use the point-slope form of a line.
The equation of a line in point-slope form is given by:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line and m is the slope of the line.
In this case, the point (x1, y1) on the tangent line is (3, f(3)) = (3, 2). The slope of the tangent line is f'(3) = 5.
Plugging these values into the point-slope form equation, we have:
y - 2 = 5(x - 3)
Expanding and simplifying:
y - 2 = 5x - 15
Now, if we want to find the x-coordinate where y = 0 (the zero of f), we can set y = 0:
0 - 2 = 5x - 15
2 = 5x - 15
Adding 15 to both sides:
17 = 5x
Dividing both sides by 5:
x β 3.4
So, using the tangent line at x = 3, an approximation to a zero of f is x β 3.4.