The function g is given by the formula
g(x)= ∫[0,x/2] e^(-t^2)dt. An equation for the tangent line to the graph of g at the point x=1 is
a) y-0.461=0.779(x-1)
b) y-0.461=0.389(x-1)
c) y-0.461=1.558(x-1)
d) y-0.461=0.184(x-1)
e) y-0.461=0.368(x-1)
I took the derivative and got e^(-x^2)*(1/2 ) in order to find the slope and then plugged in 1 but i am keep getting 0.184 which is wrong. Please help
You have to take f(x/2), not f(x). Thus, the derivative is
1/2 e^(-(x/2)^2) = 1/2 e^(-x^2/4)
So, at x=1, the slope is 1/2 e^(-1/4) = 0.3894
So, it looks like (b) is the choice. See the graphs here:
http://www.wolframalpha.com/input/?i=plot+y%3D%E2%88%9A%CF%80%2F2+*+erf%28x%2F2%29%2C+y%3D0.3894%28x-1%29+%2B0.461
To find the equation for the tangent line to the graph of g at the point x = 1, we need to find both the slope and the y-intercept of the tangent line.
First, let's find the slope. To do this, we need to take the derivative of g(x). Given that g(x) = ∫[0,x/2] e^(-t^2)dt, we can use the Fundamental Theorem of Calculus to differentiate the integral with respect to x. Applying the chain rule, we have:
g'(x) = d/dx ∫[0,x/2] e^(-t^2)dt = e^(-((x/2)^2)) * (1/2) = e^(-(x^2)/4) * (1/2).
To find the slope at x = 1, we substitute x = 1 into the derivative:
g'(1) = e^(-(1^2)/4) * (1/2) = e^(-1/4) * (1/2).
Next, let's find the y-coordinate of the point where the tangent line intersects the graph of g at x = 1. We can find this by evaluating g(1):
g(1) = ∫[0,1/2] e^(-t^2)dt.
Now, we need to evaluate this integral. Unfortunately, there is no standard elementary function for the anti-derivative of e^(-t^2), so we have to use numerical methods to approximate the value. One common method is to use numerical integration techniques, such as Simpson's rule or the midpoint rule. However, without the specific bounds of the integral, it is not possible to provide an exact value for g(1).
Therefore, based on the information given, we can only find the slope of the tangent line at x = 1, which is g'(1) = e^(-1/4) * (1/2). Since none of the answer choices include this value, it seems that the options provided may not be correct.
To find the equation for the tangent line to the graph of g at the point x=1, you have correctly taken the derivative of g(x) with respect to x. However, it seems there might be a mistake in your calculation.
Let's go through the process step by step:
1. Start with the function g(x) = ∫[0,x/2] e^(-t^2)dt.
2. To find the derivative of g(x), we will use the Fundamental Theorem of Calculus. The derivative of the integral of a function is simply the original function evaluated at the upper limit of integration.
3. Take the derivative of the function e^(-t^2) with respect to t: d/dt(e^(-t^2)) = -2te^(-t^2).
4. Now, evaluate the derivative at the upper limit of integration, which is x/2:
g'(x) = -2(x/2)e^(-(x/2)^2) = -xe^(-(x/2)^2).
5. To find the slope of the tangent line at x=1, substitute x=1 into the derivative:
g'(1) = -(1)e(-(1/2)^2) = -e^(-1/4).
So, the correct slope of the tangent line at x=1 is -e^(-1/4).
Using the point-slope form of the equation of a line (y - y1) = m(x - x1), where m is the slope and (x1, y1) is a point on the line (in this case, (x1, g(x1))), we can proceed.
We already have the slope m = -e^(-1/4) and the point (x1, g(x1)) = (1, g(1)).
To find g(1), we substitute x=1 into the original function g(x):
g(1) = ∫[0,1/2] e^(-t^2)dt.
Now, we need to evaluate this integral. This can be done either analytically or numerically using methods like numerical integration or computer software.
Once you find the value of g(1), you can use it along with the slope to determine the correct equation for the tangent line by substituting the values in the point-slope form mentioned earlier. The correct answer option will be the one that matches the equation obtained.
Remember, taking the integral of e^(-t^2) is not trivial, and there is no elementary function to express it. So, you may need to use numerical methods or software to find the definite integral.
Hope this helps!