Find the slope of the functions graph at a given point. Then find an equation for the line tangent to the graph there.
f(x)=2*square root of x. (1,2)
Could someone please help me figure this one out?
given
f(x)=2√x
Therefore f(1)=2*√1=2, which means that (x0,y0)=(1,2) is a point on the curve (given).
Find f'(x)=(d/dx)2√x=1/√x
the tangent at (1,2) is the line passing through (1,2) with a slope equal to it's slope, namely f'(1).
Thus the required line is:
L : (y-y0)=f'(1)*(x-x0)
or
L : (y-2)=f'(1)*(x-1)
I will let you evaluate f'(1) to get the final solution.
To find the slope of the graph of a function at a given point, we can use the concept of derivatives. The derivative of a function gives us the slope of the tangent line to the graph at any point.
Let's find the derivative of the function f(x)=2√x using the power rule for derivatives. The power rule states that if we have a function of the form f(x) = ax^n, then its derivative is given by f'(x) = nax^(n-1).
In our case, f(x) = 2√x can be written as f(x) = 2x^(1/2). Applying the power rule, we get f'(x) = (1/2) * 2 * x^(1/2 - 1) = x^(-1/2) = 1/√x.
Now, we can find the slope of the graph at the point (1,2) by substituting x=1 into the derivative:
f'(1) = 1/√1 = 1/1 = 1.
So, the slope of the graph at the point (1,2) is 1.
To find an equation for the line tangent to the graph at this point, we can use the point-slope form of a linear equation, which is given by y - y1 = m(x - x1), where (x1, y1) is the point and m is the slope.
Using the point (1,2) and slope 1, the equation for the tangent line becomes:
y - 2 = 1(x - 1) (simplifying)
y - 2 = x - 1
y = x + 1
Therefore, the equation for the line tangent to the graph of f(x) = 2√x at the point (1,2) is y = x + 1.