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 slope of the tangent line to the graph of f at x is equal to f′(x). So, to find the x value at which the tangent line has slope 2, we need to solve the equation f′(x) = 2.
f′(x) = 0.1x + e^0.25x
Setting f′(x) equal to 2, we get:
2 = 0.1x + e^0.25x
We can’t solve this equation algebraically, so we’ll need to use numerical methods. One common method is the Newton-Raphson method, which involves making an initial guess for the solution and then refining the guess using the formula:
x1 = x0 – f(x0)/f′(x0)
where x0 is the initial guess and x1 is the refined guess.
Let’s make an initial guess of x0 = 2. Plugging this into the formula, we get:
x1 = 2 – (0.1(2) + e^(0.25(2)) – 2)/(0.1 + 0.25e^(0.25(2)))
Evaluating this expression gives:
x1 ≈ 3.26
This is our refined guess. We can continue the process to get a more accurate answer. For example, plugging x1 back into the formula gives:
x2 = 3.26 – (0.1(3.26) + e^(0.25(3.26)) – 2)/(0.1 + 0.25e^(0.25(3.26)))
Evaluating this expression gives:
x2 ≈ 3.20
We can continue this process until we reach a desired level of accuracy. For example, if we keep going until the values of x stop changing after the second decimal place, we get:
x ≈ 3.19
So the value of x at which the tangent line to the graph of f has slope 2 is approximately 3.19.
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