Question

if ( f(x) = 7x^2 - x^3 ), find ( f(1) ) and use it to find an equation of the tangent line to the curve ( y = 7x^2 - x^3 ) at the point ( (1, 6) ). 8. find ( f(a) ) where ( f(t) = \frac{2t + 2}{t + 3} ).

Explanation:

First Problem (Finding \( f'(1) \) and tangent line)

Step1: Find the derivative of \( f(x) \)

Using the power rule, if \( f(x) = 7x^2 - x^3 \), then \( f'(x)=\frac{d}{dx}(7x^2)-\frac{d}{dx}(x^3)=14x - 3x^2 \).

Step2: Evaluate \( f'(1) \)

Substitute \( x = 1 \) into \( f'(x) \): \( f'(1)=14(1)-3(1)^2=14 - 3 = 11 \).

Step3: Find the tangent line equation

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(1,6) \) and \( m = f'(1)=11 \).
Substitute the values: \( y - 6 = 11(x - 1) \).
Simplify: \( y-6=11x - 11 \), so \( y = 11x-5 \).

Step1: Use the quotient rule

The quotient rule states that if \( f(t)=\frac{u(t)}{v(t)} \), then \( f'(t)=\frac{u'(t)v(t)-u(t)v'(t)}{[v(t)]^2} \).
Let \( u(t)=2t + 2 \), so \( u'(t)=2 \); let \( v(t)=t + 3 \), so \( v'(t)=1 \).

Step2: Apply the quotient rule

\( f'(t)=\frac{2(t + 3)-(2t + 2)(1)}{(t + 3)^2} \).

Step3: Simplify the numerator

Expand the numerator: \( 2t+6-2t - 2=4 \).
So \( f'(t)=\frac{4}{(t + 3)^2} \).

Step4: Evaluate at \( t = a \)

Substitute \( t=a \) into \( f'(t) \), we get \( f'(a)=\frac{4}{(a + 3)^2} \).

Answer:

(for \( f'(1) \)): \( 11 \)