Given that the graph of f(x) passes through the point (2,5) and that the slope of its tangent line at (x, f(x))is 3x+5, what is f(3) ?
if dy/dx = 3x+5 , then
y = (3/2)x^2 + 5x + c
when x=2, y = 5
5 = (3/2)(4) + 10 + c
5 = 6 + 10 + c
c = -11
f(x) = (3/2)x^2 + 5x - 11
f(3) = (3/2)(9) +15 - 11 = 35/2
Well, well, well, if the slope of the tangent line is 3x + 5, then we can integrate that to find the equation of the original function f(x). Integrating gives us f(x) = (3/2)x^2 + 5x + C, where C is a constant.
But wait, we can make use of the fact that the graph passes through the point (2,5) to find the value of C. Plug in x = 2 and f(2) = 5 into the equation and solve for C.
(3/2)(2)^2 + 5(2) + C = 5
6 + 10 + C = 5
C = -11
Ouch! Looks like our constant is -11. So the equation for f(x) is f(x) = (3/2)x^2 + 5x - 11.
Now, let's find f(3) by plugging in x = 3 into our function:
f(3) = (3/2)(3)^2 + 5(3) - 11
f(3) = 13.5 + 15 - 11
f(3) = 17.5
So, f(3) = 17.5. Voila!
To find the value of f(3), we need to determine the equation of the tangent line at the point (2,5) and then figure out where this tangent line intersects the x-axis, which will be the x-coordinate of f(3).
First, let's find the equation of the tangent line at (2,5) using the slope given. The slope-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept.
Given slope = 3x + 5, we can plug in the coordinates (2,5) to find the y-intercept:
5 = 3(2) + b
5 = 6 + b
b = -1
So the equation of the tangent line is y = 3x - 1.
Now, let's find where this tangent line intersects the x-axis. In other words, we need to find the x-coordinate when y = 0.
0 = 3x - 1
3x = 1
x = 1/3
Therefore, f(3) is equal to the x-coordinate of the intersection point, which is 1/3.
To find the value of f(3), we can start by using the given information to find the equation of the tangent line at (x, f(x)).
We know that the slope of the tangent line at (x, f(x)) is given by 3x + 5. This means that at any given point (x, f(x)), the slope of the tangent line is 3x + 5.
To find the equation of the tangent line, we can use the point-slope form of a linear equation. The 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, we can plug in the values (2, 5) for (x1, y1) and 3x + 5 for m. This gives us:
y - 5 = (3x + 5)(x - 2)
Expanding the right side of the equation, we get:
y - 5 = 3x^2 - x - 10
Next, we want to find the equation of the graph of f(x). This means that we want to find an equation that satisfies the given conditions and passes through the point (2, 5).
Since the equation of the tangent line gives us the slope at any point (x, f(x)), we can integrate 3x + 5 with respect to x to find f(x).
∫ (3x + 5) dx = ∫ 3x dx + ∫ 5 dx
This gives us:
f(x) = (3/2)x^2 + 5x + C
Where C is a constant of integration.
Now, we can use the fact that the graph passes through the point (2, 5) to find the value of C. Plugging in the values (2, 5) for (x, f(x)), we get:
5 = (3/2)(2)^2 + 5(2) + C
Simplifying, we find:
5 = 6 + 10 + C
Cancelling and rearranging, we get:
C = -11
Finally, we can substitute x = 3 into the equation for f(x) to find f(3):
f(3) = (3/2)(3)^2 + 5(3) - 11
Evaluating this expression, we find:
f(3) = 9/2 + 15 - 11
Simplifying, we get:
f(3) = 4.5 + 4
Therefore, the value of f(3) is 8.5.