Find the slope of the tangent line to the graph of the given function at the given value of x.
y=-5x^1/2+x^3/2; x=25
just evaluate f'(x) at x=25
f'(x) = -5/2 x^(-1/2) + 3/2 x^(1/2)
= (-5/2)(1/5) + 3/2 (5)
= 7
To find the slope of the tangent line to the graph of the function at the given value of x, we need to find the derivative of the function and then evaluate it at x=25.
First, let's find the derivative of the function y=-5x^(1/2)+x^(3/2) using the power rule:
The power rule states that for any real number n, if f(x) = x^n, then f'(x) = n*x^(n-1).
So, applying the power rule to each term of the function, we have:
y' = d/dx (-5x^(1/2)) + d/dx (x^(3/2))
= -5*(1/2)*x^(-1/2) + (3/2)*x^(1/2)
Simplifying further, we get:
y' = -5/2*x^(-1/2) + 3/2*x^(1/2)
Now, let's substitute x=25 into the derivative to find the slope of the tangent line at x=25:
Slope at x=25 (m) = y'(25)
= -5/2*(25)^(-1/2) + 3/2*(25)^(1/2)
To evaluate this expression, let's simplify the radicals:
(25)^(-1/2) = 1/(sqrt(25)) = 1/5
(25)^(1/2) = sqrt(25) = 5
So, substituting these values into the expression for the slope, we have:
m = -5/2*(1/5) + 3/2*(5)
= -5/10 + 15/2
= -1/2 + 15/2
= 14/2
= 7
Therefore, the slope of the tangent line to the graph of the function at x=25 is 7.
To find the slope of the tangent line to the graph of the function at a given value of x, you can use calculus and find the derivative of the function with respect to x. The derivative gives you the rate of change of the function at any given point.
Let's find the derivative of the given function:
y = -5x^(1/2) + x^(3/2)
To find the derivative, we differentiate each term of the function separately.
For the first term, -5x^(1/2), the power rule of differentiation tells us that the derivative is:
(d/dx) (-5x^(1/2)) = -5 * (1/2) * x^((1/2)-1) = -5 * (1/2) * x^(-1/2) = -5/2 * x^(-1/2)
For the second term, x^(3/2), we have:
(d/dx) (x^(3/2)) = (3/2) * x^((3/2)-1) = (3/2) * x^(1/2)
Now, let's find the derivative of the entire function by adding the derivatives of each term:
dy/dx = (-5/2) * x^(-1/2) + (3/2) * x^(1/2)
Now, to find the slope of the tangent line at x = 25, we substitute x = 25 into the derivative:
dy/dx = (-5/2) * 25^(-1/2) + (3/2) * 25^(1/2)
Simplifying this expression gives:
dy/dx = (-5/2) * 1/√25 + (3/2) * √25 = (-5/2) * 1/5 + (3/2) * 5 = -5/10 + 15/10 = 10/10 = 1
Therefore, the slope of the tangent line to the graph of the function at x = 25 is 1.