Answer:
1) [tex]f'(x)= \frac{3}{2\sqrt{3x-5}}[/tex]
2) [tex]y=\frac{3}{4}x-\frac{1}{4}[/tex]
Step-by-step explanation:
So we have the function:
1)
And we want to find its derivative using the limit definition:
[tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]
So, let's do so:
[tex]\lim_{h \to 0} \frac{\sqrt{3(x+h)-5}-\sqrt{3x-5}}{h}[/tex]
Simplify the first square root:
[tex]\lim_{h \to 0} \frac{\sqrt{3x+3h-5}-\sqrt{3x-5}}{h}[/tex]
Now, let's remove the square root by multiplying by its conjugate. So:
[tex]\lim_{h \to 0} \frac{\sqrt{3x+3h-5}-\sqrt{3x-5}}{h}\cdot (\frac{ \sqrt{3x+3h-5}+\sqrt{3x-5} }{\sqrt{3x+3h-5}+\sqrt{3x-5} })[/tex]
In the numerator, difference of two squares. In the denominator, multiply:
[tex]\lim_{h \to 0} \frac{(3x+3h-5)-(3x-5)}{h(\sqrt{3x+3h-5}+\sqrt{3x-5})}[/tex]
Distribute the numerator:
[tex]\lim_{h \to 0} \frac{(3x+3h-5)-3x+5}{h(\sqrt{3x+3h-5}+\sqrt{3x-5})}[/tex]
Simplify:
[tex]\lim_{h \to 0} \frac{3h}{h(\sqrt{3x+3h-5}+\sqrt{3x-5})}[/tex]
Cancel out the h:
[tex]\lim_{h \to 0} \frac{3}{\sqrt{3x+3h-5}+\sqrt{3x-5}}[/tex]
Direct substitution:
[tex]= \frac{3}{\sqrt{3x+3(0)-5}+\sqrt{3x-5}}[/tex]
Simplify:
[tex]= \frac{3}{\sqrt{3x-5}+\sqrt{3x-5}}[/tex]
Combine like terms:
[tex]= \frac{3}{2\sqrt{3x-5}}[/tex]
Therefore:
[tex]f'(x)= \frac{3}{2\sqrt{3x-5}}[/tex]
Your answer is indeed correct!
2)
Now, let's find the equation of the tangent line to the graph at x=3.
Remember what the derivative gives us. The derivative tells us the slope of the tangent line to a graph at a certain point. So, let's substitute 3 into f'(x) to find the slope of our tangent line:
[tex]f'(3)= \frac{3}{2\sqrt{3(3)-5}}[/tex]
Multiply and subtract:
[tex]f'(3)= \frac{3}{2\sqrt{9-5}}\\f'(3)= \frac{3}{2\sqrt{4}}[/tex]
Simplify:
[tex]f'(3)= \frac{3}{2(2)}\\f'(3)= \frac{3}{4}[/tex]
Therefore, the slope of our tangent line is 3/4.
Now, let's find the equation of the line using the point-slope form. So, we need to point at x=3 of the original graph. So, substitute 3 into the original function:
[tex]f(3)=\sqrt{3(3)-5}[/tex]
Multiply, subtract, and simplify:
[tex]f(3)=\sqrt{9-5}\\f(3)=\sqrt{4}\\f(3)=2[/tex]
Therefore, our point is (3,2).
Now, we can use the point-slope form, where:
[tex]y-y_1=m(x-x_1)[/tex]
Let (3,2) be (x₁, y₁) and substitute 3/4 for m. Therefore:
[tex]y-2=\frac{3}{4}(x-3)[/tex]
Distribute:
[tex]y-2=\frac{3}{4}x-\frac{9}{4}[/tex]
Add 2 to both sides:
[tex]y=\frac{3}{4}x-\frac{1}{4}[/tex]
And we're done!
