can someone explain to me how this question is done?
Find an equation for the line tangent to the graph of the given function at the indicated point.
f(x) = x3 – x2 at (0, 0)
f(x) = x3 – x2 at (0, 0)
To find the value for f(x), substitute, the x value in the original equation.
f(0) = 0^3 - 0^2
f(0) = 0 - 0
f(0) = 0
Now, take the first derivative of the original f(x) equation to get the slope of the tangent line.
Use the power rule, which states that:
If f(x) = x^n, then f'(x) = nx^(n-1).
So, f'(x) = 3*x^2 - 2*x
To find the slope substitute the x value into f'(x):
f'(0) = 3*0^2 - 2*0
f'(0) = 0 - 0
f'(0) = 0
So now we can set up the equation of the tangent line:
y = mx + b. Note, f'(x) = m and f(x) = y.
So, 0 = 0*0 + b
b = 0.
So the equation of the tangent line of f(x) at (0,0) is written as
y = 0
I hope this is useful.
y = 6 and u = 8x - 5
u^2
Find expression of the form dy/dx
Is this y = 6/(u^2)?
First, rewrite y as: y = 6*(u^-2). It is the same as the original y equation.
Now, we differentiate. The right side is with respect to y and the left side is with respect to u:
dy = 6*-2*(u^-3)du.
This was obtained using the power rule, which states that if y = x^n, then dy/dx = n*x^n-1
So, we have dy = 6*-2*(u^-3)du. We're halfway done since we have to have it in terms of dy/dx.
Since u is denoted above as u = 8x - 5, we should differentiate this equation. The left side is with respect to u and the right side is with respect to x:
du = 8 dx.
Now, substitute this into the dy equation above along with the u value to get:
dy = 6*-2*((8x-5)^-3)*8dx
Divide both sides by dx and simplify to yield:
dy/dx = -96*(8x-5)^-3.
It wouldn't be a bad idea to review the power rule and the chain rule since both are involved in this problem.
Am currently doing a practice sheet but cant do half of it, need some help please. Here goes
Determine the continuity of the function at the given points.
f(x) = -1, for x =-2
0.5, for x not equal to -2
at x = -2 and x = -3
Answer the question.
Is the function given by
f(x) =x+4/ x^2 -9x+8
continous at x = 1? Why or why not?
Find the intervals on which the function is continuous.
Is the function given by
f(x) = 1/ x+1^2 +2
continuous on ℛ? Why or why not?
Find a simplified difference quotient for the function.
f(x) = 9^x3
Find a simplified form of the difference quotient for the function.
f(x) = square root of x-4
Calculate the requested derivative from the given information.
Given f(u) = u-1/u+1, g(x) = u = square root of x, find (f ∘ g)'( 4).