Find the tangent line to f(x)=7x+4ex at x=0
Find the tngent line to f(x)=3x^2ln x at x=1
Find the tangent line to f(x) = ln(x) log2(x) at x=2
Well, I will start one of them
tangent line to f(x)=7x+4e^x at x=0 I assume you mean
slope = dy/dx = m = 7 + 4 e^x
when x = 0
slope = m = 7 + 4 = 11
so
y = 11 x + b
when x = 0, y = 4
so
4 = 11* (0) + b
so b = 4
y = 11 x + 4
1.
I will interpret that as f(x)=7x+4e^x at x=0
f'(x) = 7 + 4e^x
f'(0) = 7 + 4 = 11
f(0) = 0+4 = 4
equation of tangent:
(y-4) = 11(x-0)
arrange to whatever form you need
2.
f(x)= (3x^2)(ln x) at x=1
use product rule and follow steps from above
3.
I will interpret your log2(x) as log2 x
recall that log2 x
= lnx / ln2 or (1/ln2) lnx
so now you have another product rule like in #2
#3. No product rule needed, since ln2 is just a constant.
To find the tangent line to a function at a specific point, you need to find the slope of the tangent line and the point where the line intersects the function. The slope of the tangent line is equal to the derivative of the function evaluated at that point.
For the first function, f(x) = 7x + 4e^x, we need to find the derivative. The derivative of 7x is 7, and the derivative of 4e^x is 4e^x since the derivative of e^x is itself. Therefore, the derivative of f(x) is 7 + 4e^x.
To find the slope of the tangent line at x=0, we need to evaluate the derivative at that point. Substituting x=0 into the derivative expression, we get 7 + 4e^0 = 7 + 4 = 11. So the slope of the tangent line is 11 at x=0.
We also need to find the point where the tangent line intersects the function. At x=0, f(x) = 7(0) + 4e^(0) = 0 + 4 = 4. So the point of intersection is (0, 4).
Therefore, the equation of the tangent line to f(x) = 7x + 4e^(x) at x=0 is y = 11x + 4.
For the second function, f(x) = 3x^2ln(x), we need to find the derivative. The product rule is required for this calculation. The derivative of 3x^2 is 6x, and the derivative of ln(x) is 1/x. Therefore, the derivative of f(x) is 6x * ln(x) + 3x^2 * (1/x).
To find the slope of the tangent line at x=1, we substitute x=1 into the derivative expression. This gives us 6(1) * ln(1) + 3(1)^2 * (1/1) = 0 + 3 = 3. So the slope of the tangent line is 3 at x=1.
At x=1, f(x) = 3(1)^2 * ln(1) = 3(0) = 0. So the point of intersection is (1, 0).
Therefore, the equation of the tangent line to f(x) = 3x^2ln(x) at x=1 is y = 3(x - 1).
For the third function, f(x) = ln(x) * log2(x), we need to find the derivative. The product rule is required again. The derivative of ln(x) is 1/x, and the derivative of log2(x) is 1/(x * ln(2)). Therefore, the derivative of f(x) is (1/x) * log2(x) + ln(x) * (1/(x * ln(2))).
Substituting x=2 into the derivative expression, we get (1/2) * log2(2) + ln(2) * (1/(2 * ln(2))). This simplifies to (1/2) * 1 + ln(2) * (1/(2 * ln(2))). The ln(2) terms cancel out, and we are left with (1/2) + (1/2) = 1. So the slope of the tangent line is 1 at x=2.
At x=2, f(x) = ln(2) * log2(2) = ln(2) * 1 = ln(2). So the point of intersection is (2, ln(2)).
Therefore, the equation of the tangent line to f(x) = ln(x) * log2(x) at x=2 is y = x - (2 - ln(2)).
I hope this explanation helps you understand how to find tangent lines to functions at specific points.