Derivative question from Berkeley Problems (Real Analysis)
Posted: Thu Sep 15, 2011 7:07 pm
In the solution to problem 1.1.11, there's is one line in the solution that is giving me trouble. Here is the problem:
Suppose that $$f$$ is a twice differentiable real function such that $$f''(x)>0$$ for all $$x\in[a,b]$$. Find all real numbers $$c$$ at which the area between the graph $$y=f(x)$$, the tangent to the graph at $$(c,f(c))$$, and the lines $$x=a$$, $$x=b$$ attains its minimum value.
At some point we need to differentiate (with respect to c)
$$A(c) = \int_a^b ( f(x) - f(c) - f'(c)(x-c) ) dx$$
and get
$$A'(c) = -f''(c) \int_a^b (x-c) dx$$,
but I'm getting other things which don't allow successful solution to the problem (my expression for $$A'(c)$$ doesn't have a $$c$$ in it).
Can someone explain?
Suppose that $$f$$ is a twice differentiable real function such that $$f''(x)>0$$ for all $$x\in[a,b]$$. Find all real numbers $$c$$ at which the area between the graph $$y=f(x)$$, the tangent to the graph at $$(c,f(c))$$, and the lines $$x=a$$, $$x=b$$ attains its minimum value.
At some point we need to differentiate (with respect to c)
$$A(c) = \int_a^b ( f(x) - f(c) - f'(c)(x-c) ) dx$$
and get
$$A'(c) = -f''(c) \int_a^b (x-c) dx$$,
but I'm getting other things which don't allow successful solution to the problem (my expression for $$A'(c)$$ doesn't have a $$c$$ in it).
Can someone explain?