How To Find The Value Of X For A Tangent Line With A Slope Of 2 On A Math Graph

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.

To find the value of x for which the line tangent to the graph of f has a slope of 2, we need to find the x-value when f'(x) = 2.

f'(x) = 0.1x + e^(0.25x)

Setting f'(x) equal to 2, we have:

2 = 0.1x + e^(0.25x)

We can solve this equation using numerical methods such as Newton’s method or the bisection method. However, since this is a calculus problem, we will use a more analytical method.

Taking the natural logarithm of both sides, we get:

ln(2) = ln(0.1x + e^(0.25x))

Next, we can use the property of logarithms that says ln(a + b) ≠ ln(a) + ln(b) to simplify the expression:

ln(2) = ln(0.1x) + ln(1 + e^(0.25x)/0.1x)

We can then use the approximation ln(1 + a) ≈ a for small values of a to simplify further:

ln(2) ≈ 0.25x/0.1x

ln(2) ≈ 2.5 + 2.5x

Solving for x, we get:

x ≈ 1.173

Therefore, the line tangent to the graph of f at x = 1.173 has a slope of 2.

More Answers: