Evaluate the approximate tangent slope to f(x)=x^3/8 when x=4
If you mean f(x)=x^(3/8), then the slope at any x is
f'(x) = 3/8 x^(-5/8)
So, when x=4, f'(4) = 3/8 * 4^(-5/8) = 0.157
No. Sorry it's actually f(x)=(x^3)/8 when x=4
d/dx (1/8) x^3 = (1/8) d/dx x^3
= (1/8) (3 x^2)
= (3/8) x^2
when x = 4
= (3/8)(16)
= 6
AT last someone who meant what she wrote, rather than just being careless with parentheses.
However, I'm surprised that given my solution, you did not just fix my function and apply the same logic.
Thanks both of you!!!
To evaluate the approximate tangent slope to the function f(x) = x^3/8 when x = 4, we can use the concept of derivatives. The derivative of a function gives us the slope of the tangent line at any point on the graph.
To find the derivative of f(x), we can apply the power rule. For any function of the form f(x) = x^n, the derivative is given by f'(x) = n*x^(n-1).
Applying the power rule to f(x) = x^3/8, we have:
f'(x) = (3/8) * x^(3/8 - 1) = (3/8) * x^(-5/8)
Now, to determine the approximate tangent slope at x = 4, we substitute this value into the derivative:
f'(4) = (3/8) * 4^(-5/8)
Simplifying:
f'(4) = (3/8) * (1/∛2)
To approximate this value, we can use a calculator or mathematical software.
Therefore, the approximate tangent slope to f(x) = x^3/8 when x = 4 is (3/8) * (1/∛2).