Find k such that the line is tangent to the graph of the function. Function: f(x)=k�ã(x)
Line: y=x+4
I will assume your equation is
f(x) = k√x = k x^(1/2)
then f'(x) = (k/2)x^(-1/2)
but the slope of y = x+4 is 1
so (k/2)x^(-1/2) = 1
k/(2√x) = 1
k = 2√x
then y = 2√x √x = 2x
sub back into y = x+4
2x = x+4
x = 4
so k= 2√4 = 4
y=2x²√ 2-x
To find the value of 'k' such that the line is tangent to the graph of the function, we need to determine the point of tangency.
The graph of the function is expressed as f(x) = k�ã(x), and the equation of the line is given as y = x + 4.
For a line to be tangent to a curve at a specific point, the slope of the line should be equal to the slope of the curve at that point. Therefore, we need to find the slope of both the line and the curve.
The slope of the line y = x + 4 is 1, as the coefficient of 'x' is 1.
To find the slope of the function f(x) = k�ã(x), we need to differentiate it. The derivative will give us the slope at any given point.
Differentiating f(x) = k�ã(x) with respect to 'x', we get:
f'(x) = k * d/dx (x�ã)
The derivative of x�ã can be found using the power rule of differentiation. For any real number 'n', d/dx (x^n) = n * x^(n-1).
Applying the power rule, we get:
f'(x) = k * (1/2) * x^((1/2)-1)
= k * (1/2) * x^(-1/2)
= (k/2) * (1/x^(1/2))
Now, to find the slope at the point of tangency, we set the derivative equal to the slope of the line:
(k/2) * (1/x^(1/2)) = 1
Now, we solve this equation for 'k' by isolating it:
(k/2) = x^(1/2)
k = 2 * x^(1/2)
k = 2 * sqrt(x)
Since the line is tangent to the graph of the function, it should only intersect at one point. Therefore, the value of 'x' can be found by equating the y-values of the line and the function:
x + 4 = k�ã(x)
Substituting the expression for 'k', we have:
x + 4 = 2 * sqrt(x) * �ã(x)
Now, solve this equation to find the value of 'x'. Once 'x' is determined, substitute it back into the expression for 'k' to find the exact value of 'k' such that the line is tangent to the graph of the function.