The derivative of a function f is given by f′(x)=0.1x+e^0.25x. At what value of x for x>0 does the line tangent to the graph of f at x have slope 2 ?
2.287Since the derivative at a point is the slope of the line tangent to the graph at that point, the calculator is used to solve f′(x)=0.1x+e^(0.25x)=2.
We know that the derivative of a function at a given point gives us the slope of the tangent line to the graph of the function at that point. So, if we want to find the value of x for which the tangent line has a slope of 2, we need to find the point on the graph where the derivative equals 2.
Let’s set up the equation:
f′(x)=0.1x+e^0.25x=2
We can simplify this equation by subtracting 2 from both sides:
0.1x+e^0.25x-2=0
Unfortunately, there is no analytical solution to this equation. However, we can use numerical methods to approximate the solution.
One possible method is to use Newton’s method, which involves iteratively refining an initial guess until we reach a solution. Let’s start with an initial guess of x=3:
x1=3 – (0.1(3) + e^(0.25(3)) – 2)/(0.1 + 0.25 e^(0.25(3))) = 2.714
x2=2.714 – (0.1(2.714) + e^(0.25(2.714)) – 2)/(0.1 + 0.25 e^(0.25(2.714))) = 2.683
x3=2.683 – (0.1(2.683) + e^(0.25(2.683)) – 2)/(0.1 + 0.25 e^(0.25(2.683))) = 2.682
We can continue this process until we reach a desired level of accuracy. In this case, we can see that the value of x approaches 2.682 as we iterate. So, the line tangent to the graph of f at x=2.682 has slope 2.
Therefore, the value of x for x>0 where the line tangent to the graph of f at x has slope 2 is approximately 2.682.
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